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Solutions to homework problems

November 25, 2015

Contents

1 Homework Assignment # 1 1

2 Homework assignment #2 6







9 Homework Assignment #9 41


1 Homework Assignment # 1

1. Show that the ε-δ definition of continuity of a map f : X → Y between metric spaces isequivalent to the definition of continuity in terms of open subsets.

Proof. Suppose f is ε-δ-continuous (i.e., continuous in the sense of the ε-δ definition ofcontinuity) and that U is an open subset of Y . We need to show that f−1(U) is an opensubset of X. The definition of an open subset U of a metric space implies that U can bewritten as a union of open balls Bi, i ∈ I. Then

f−1(U) = f−1(⋃i∈I

Bi) =⋃i∈I

f−1(Bi),

and hence it suffices to show that the inverse image of a ball B in Y is an open subset ofX. So let B = Bε(y0) = {y ∈ Y | d(y, y0) < ε} and let x0 be an element of f−1(B). Then

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the continuity of f in the ε-δ sense implies that there is some δ > 0 such that Bδ(x0) ⊂f−1(Bε(y0)). This shows that f is continuous.

Conversely, assume that f is continuous (as defined via open sets). To show that f isε-δ-continuous let x ∈ X and ε > 0. Then f−1(Bε(f(x)) is an open subset of X containingx. Hence there is some δ > 0 such that the ball Bδ(x) is contained in f−1(Bε(f(x))). Inother words, f(Bδ(x)) ⊂ Bε(f(x)), which implies

∀y ∈ X d(y, x) < δ ⇒ d(f(y), f(x)) < ε

2. a) Suppose T, T′ are topologies on a set X which are generated by bases B and B′,respectively. Show that the topologies agree if and only if the bases B, B′ are equivalent inthe sense that every B ∈ B is the unions of subsets belonging to B′ and every B′ ∈ B′ is theunion of subsets belonging to B.

b) Show that the metric topology on Rn determined by the Euclidean metric is equal tothe metric topology determined by the `1-metric d1(x, y) =

∑ni=1 |xi − yi|.

c) Show that the product topology on Rm × Rn (with each factor equipped with themetric topology) agrees with the metric topology on Rm+n = Rm × Rn. Hint: Use thefreedom provided by part (b) to choose whether to work with the Euclidean metric or the`1-metric to make this easier.

Proof. Part (a). Suppose U ∈ T, i.e., U is an open subset of X w.r.t. the topology T. ThenU is the union of subsets Bi belonging to B. By assumption, each Bi is the union of subsetsbelonging to B′, and hence U is the union of subsets belonging to B′ and consequentlyU ∈ T′. Analogously, U ∈ T′ implies U ∈ T, and hence the topologies T and T′ coincide.

Part (b). We recall that the metric topology on a set X equipped with a metric d(x, y) isgenerated by the collection of subsets of X consisting of the open balls Bε(x) = {y ∈ X |d(y, x) < ε} of any radius ε > 0 around any point x ∈ X. Specializing to X = Rn, let usdenote by Bε(x) (resp. B′ε(x)) the open ball of radius ε around x ∈ Rn for the Euclideanmetric d (resp. the `1-metric d1), and by B (resp. B′) the collection of all balls Bε(x) (resp.B′ε(x)). By part (a) it suffices to show that

1. every ball Bε(x) is a union of balls B′δ(y) ⊂ Bε(x), and

2. every ball B′ε(x) is a union of balls Bδ(y) ⊂ B′ε(x).

Comparing balls defined w.r.t. the Euclidean metric d with those w.r.t. the metric d1 willrequire inequalities comparing the distances d(x, y) and d1(x, y) for x, y ∈ Rn. We recallthat d(x, y) = ||x− y||, and d1(x, y) = ||x− y||1, where

||v|| :=√v2

1 + · · ·+ v2n and ||v||1 := |v1|+ · · ·+ |vn|

is the standard norm (resp. the 1-norm) of a vector v ∈ Rn. These norms are related by theinequalities

||v|| ≤ ||v||1 ||v||1 ≤ n||v||.

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The first inequality follows from

||v||2 = |v1|2 + · · ·+ |vn|2 ≤ (|v1|+ · · ·+ |vn|)(|v1|+ · · ·+ |vn|) = ||v||21.

The second from

||v||1 = |v1|+ · · ·+ |vn| ≤ n max1≤i≤n

|vi| = n√

max1≤i≤n

|vi|2 ≤ n√|v1|2 + · · ·+ |vn|2 = n||v||.

We note that these inequalities imply

B′ε(x) ⊆ Bε(x) and Bε/n(x) ⊆ B′ε(x).

In particular, for y ∈ Bε(x) and δ := ε− d(y, x) > 0, we have

B′δ(y) ⊆ Bδ(y) ⊆ Bε(x). (1.1)

Here the second inclusion is a consequence of the triangle inequality for the metric d, sinceif z ∈ Bδ(y), then

d(z, x) ≤ d(z, y) + d(y, x) < δ + d(y, x) = ε,

and hence z ∈ Bε(x). This proves (1). Statement (2) is proved similarly: assume y ∈ B′ε(x)and set δ := ε− d(y, x) > 0. Then

Bδ/n(y) ⊆ B′δ(y) ⊆ B′ε(x),

where the second inclusion is a consequence of the triangle inequality for the metric d1.

Part (c). For (x, y) ∈ Rm×Rn = Rm+n, the relationship between ||x||1, ||y||1 and ||(x, y)||1is given by

||(x, y)||1 = |x1|+ · · ·+ |xm|+ |y1|+ · · ·+ |yn| = ||(x, y)||1,

which is simpler than the corresponding relationship between the Euclidean norms of thesevectors. This equality implies in particular

B′ε1(x)×B′ε2(y) ⊆ B′ε1+ε2((x, y)) and B′ε((x, y)) ⊆ B′ε(x)×B′ε(y). (1.2)

This motivates us to compare the d1 metric topology on Rm+n with the product topology ofRm and Rn, with each factor equipped with the d1 metric topology. A basis for the metrictopology for Rm+n is given by the balls B′ε((x, y)) ⊂ Rm+n. A basis for the product topologyis given by U × V , where U ⊂ Rm, V ⊂ Rn are subsets which are unions of open balls inRm and Rn, respectively. In particular, any such product U × V is a union of subsets of theform B′ε1(x)× B′ε2(y) which shows that these products of balls form a basis for the producttopology. As in part (b), the inclusions (1.2) imply that every open ball B′ε((x, y)) is a unionof product of balls, and that every product of balls is a union of balls. By part (a) thisimplies that the topologies they generate agree.

3. Let X, Y1, Y2 be topological spaces. Then the projection maps pi : Y1 × Y2 → Yi arecontinuous and a map f : X → Y1 × Y2 is continuous if and only if the component mapsfi = pi ◦ f are continuous for i = 1, 2.

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Proof. If U1 is an open subset of Y1 then p−11 (U1) = U1×Y2 is an open subset of the Cartesian

product Y1 × Y2 and hence p1 is continuous; similarly for p2.Now assume that f : X → Y1×Y2 is continuous. Then the i-th component map fi = pi◦f

is continuous as a composition of the continuous maps pi and f .Conversely, let us assume that the component maps f1 and f2 are continuous. To show

that f is continuous, it suffices to show that f−1(B) is open for an open subset B ⊂ Y1× Y2

belonging to the basis B for the product topology consisting of products B = U1 × U2 ofopen sets Ui ⊂ Yi (since any open set is of the form U =

⋃Bi for Bi ∈ B and hence

f−1(U) =⋃f−1(Bi) which is an open subset of X, provided the subsets f−1(Bi) are open).

Nowf−1(U × V ) = f−1

1 (U) ∩ f−12 (V )

is the intersection of open subsets of X (since fi is continuous and Ui is an open subset ofYi) and hence open.

4. Show that the map f : GLn(R)→ GLn(R), A 7→ A−1 is continuous (here GLn(R) ⊂ Rn2

is equipped with the metric topology).

Proof. We recall from a lemma in class that f is continuous if and only if the composition

GLn(R)f−→ GLn(R)

i↪→ Rn2

is continuous, where i is the inclusion map. Moreover, this composition is continuous if andonly if all component maps pij ◦ i ◦ f are continuous (thinking of Rn2

as the space of alln × n-matrices, pij for 1 ≤ i, j ≤ n is the projection map that takes a matrix M ∈ Rn2

toits component Mij ∈ R). We conclude that it suffices to show that for 1 ≤ i, j ≤ n the mapGLn(R)→ R, A 7→ (A−1)ij is continuous.

We recall from linear algebra that the inverse of A can be calculated by the formula

A−1 =Ct

det(A),

where det(A) is the determinant of A, and Ct is the transpose of the n× n matrix C whoseentry Cij is (−1)i+j times the (i, j)-minor of A (the determinant of the (n − 1) × (n − 1)matrix that results from deleting row i and column j of A).

This shows that each matrix entry (A−1)ij is of the form p(A)q(A)

, where p(A) and q(A) are

polynomial functions of the matrix entries of A. In particular the functions p, q : GLn(R)→ Rare continuous. Since q(A) = det(A) in non-zero for all A ∈ GLn(R), we conclude with a

lemma from class that the function p(A)q(A)

is continous.

5. Consider the following topological spaces

• The subspace T1 := {v ∈ R3 | d(v, S) = r} ⊂ R3 equipped with the subspace topology,where S = {(x, y, 0) ∈ R2 | x2 + y2 = 1} and 0 < r < 1.

• The product space T2 := S1 × S1 equipped with the product topology.

• The quotient space T3 := ([−1, 1] × [−1, 1])/ ∼ equipped with the quotient topology,where the equivalence relation is generated by (s,−1) ∼ (s, 1) and (−1, t) ∼ (1, t).

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These three spaces are homeomorphic, but we don’t yet have the tools to prove this in anefficient way. For this homework problem construct two continuous bijections between pairsof these spaces. Hint: An important part of this problem is to figure out which of thesethree spaces should be the (co)domains of the two maps you are looking for. Of course,any bijection has an inverse, but determining continuity is a different matter. Think aboutwhich type of topological space (subspace, product space, quotient space) you would preferto have as domain (resp. codomain) for a map whose continuity you want to prove.

Proof. We recall from class that checking continuity of a map f : X → Y is easy if the domainX is a quotient space and the range Y is a subspace or a product space. This suggests toconstruct continuous bijections

f : T3 → T1 and g : T3 → T2.

We first discuss the simpler map g given by the formula

g([s, t]) := ((cos πs, sin πs), (cos πt, sin πt)) ∈ S1 × S1 ⊂ R2 × R2. (1.3)

To argue that this is a well-defined bijection, we note that the map [−1, 1] → S1, s 7→(cos πs, sin πs) is surjective, and its only failure to be injective is due to the fact that it mapss = −1 and s = +1 to the same point in S1. The map above when precomposed with theprojection map [−1, 1] × [−1, 1] → T3 is the product of two copies of this map. Hence it issurjective, and its only failure of injectivity comes from points (s, t) with s = ±1 or t = ±1.This shows that g is in fact a bijection. To prove continuity of g it suffices to prove continuityof the composition

[−1, 1]× [−1, 1]pr // [−1, 1]× [−1, 1]/{±1} g // S1 × S1 i // R2 × R2 = R4 ,

where pr is the projection map, the i is the inclusion map. This map is continuous, since allits component functions, given explicitely in equation (1.3) are continuous.

Now we define the map f , first describing it geometrically. From this point of view, itwill be obvious that f is bijective. Then we will derive an explicit formula for f , which willmake it easy to argue that f is continuous.

We note that there is a surjective map p : T1 → S which sends a point v ∈ R3 to theunique point (x, y, 0) ∈ S which is closest to v. Similarly, there is a surjective map

q : T3 = ([−1, 1]× [−1, 1])/ ∼ −→ [−1, 1]/{±1} given by [s, t] 7→ [s].

We want to construct the map f such that it fits into a commutative diagram

T3 = ([−1, 1]× [−1, 1])/ ∼ f //

q

��

T1

p

��[−1, 1]/{±1} f0 // S

,

where f0 is the homeomorphism we discussed in class that sends [s] to (cos πs, sin πs, 0) ∈S ⊂ R3. We note that the fiber q−1([s]) for any fixed [s] ∈ [−1, 1]/{±1} is a circle. Similarly,for any point (x, y, 0) of the circle S ⊂ R3, the fiber p−1(x, y, 0) ⊂ T1 is a circle of radius r

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in the plane spanned by the unit vectors (x, y, 0) and (0, 0, 1) with center (x, y, 0). We notethat the map

[−1, 1]/{±1} −→R3

[t] 7→(x, y, 0) + r cosπt(x, y, 0) + r sin πt(0, 0, 1)

= ((1 + r cosπt)x, (1 + r cosπt)y, r sin πt)

is a bijection onto the circle p−1(x, y, 0). In particular, defining f on each fiber q−1([s]) tobe this map, we obtain a bijection between T3 and T1.

To argue that f is continuous, we write down f : T3 → T1 explicitly: f0 sends a point[s] ∈ [−1, 1]/{±1} to (x, y, 0) = (cos πs, sin πs, 0) and hence

f([s, t]) =((1 + r cos πt)x, (1 + r cos πt)y, r sin πt)

=((1 + r cos πt) cosπs, (1 + r cos πt) sinπs, r sin πt).(1.4)

Then f is continuous if and only if the composition

[−1, 1]× [−1, 1]pr // [−1, 1]× [−1, 1]/{±1} f // T1

i // R3

is continuous. This is continuous since its component functions are, which can be read offfrom equation (1.4).

2 Homework assignment #2

1. Show that the quotient space Dn/Sn−1 is homeomorphic to the sphere Sn.

Proof. Generalizing the case n = 1 which we did in class, we define a map

f : Dn −→ R⊕ Rn = Rn+1 by f(v) := (cos π||v||, v sin π||v||||v||

)

We recall from Calculus that t→ sinπtt

is a continuous function on R \ {0} which extends toa continuous function on all of R (by defining its value at t = 0 to be π). It follows that fis a continuous function since all its components are continuous.

It’s easy to check that f(v) belongs to Sn ⊂ Rn+1: f(0) = (0, 1) ∈ Sn, and for v 6= 0 wehave

||f(v)||2 = cos2 π||v||+ sin2 π||v||2 = 1

We conclude that the map Dn → Sn, v 7→ f(v) is continuous w.r.t. the subspace topologyon Sn since its composition with the inclusion map i : Sn → Rn+1 is the map f . Abusingnotation, we write again f : Dn → Sn.

We note that f maps any unit vector v ∈ Dn to the point (−1, 0) ∈ Rn ⊕ R. Hence finduces a well-defined map

f : Dn/Sn−1 −→ Sn [v] 7→ f(v)

This map is continuous since its composition with the projection map p : Dn → Dn/Sn−1 isthe map f which is continuous. We won’t present the details that f is bijective (if you cancome up with a map which in fact is a bijection, I believe that you could prove that it is).

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So f : Dn/Sn−1 → Sn is a continuous bijection, and hence it is a homeomorphism, sinceits range is Hausdorff (since Sn is a metric space), and its domain is compact (Dn ⊂ Rn isclosed and bouneded and hence compact by the Heine-Borel Theorem; it follows that thequotient space Dn/Sn−1 is compact).

2. Use the Heine-Borel Theorem to decide which of the topological groups GLn(R), O(n),SO(n) are compact. Provide proofs for your statements. Hint: A strategy often usefulfor proving that a subset C of Rn is closed is to show that C is of the form f−1(C ′) forsome closed subset C ′ ⊂ Rk (often C ′ consists of just one point) and some continuous mapf : Rn → Rk.

Proof. The general linear group GLn(R) ⊂ Mn×n(R) = Rn2is not compact, since it is not

bounded. To see this, consider the sequence of diagonal matrices

Ai =

i

1. . .

1

.

Considered as a vector in Rn2, the norm squared of Ai is given by

||Ai||2 = i2 + 12 + · · ·+ 12 = i2 + n− 1

which shows that GLn(R) is an unbounded subset of Rn2.

We claim that O(n) and SO(n) are compact. We begin by showing that they are boundedsubspaces of Rn2

. So let A ∈ O(n) be a matrix with column vectors v1, . . . , vn ∈ Rn. Theassumption that A belongs to O(n) means that the vi’s form an orthogonal basis of Rn.Hence

||A||2 = ||v1||2 + · · ·+ ||vn||2 = n,

which shows that O(n) and hence also SO(n) ⊂ O(n) are bounded subsets of Rn2.

To show that O(n) is a closed subset of Rn2we note that a matrix A with column vectors

vi ∈ Rn belongs to O(n) if and only if the vi’s are orthogonal, i.e., if

〈vi, vj〉 = δi,j for all 1 ≤ i, j ≤ n,

where 〈v, w〉 denotes the scalar product of vectors v, w, and the delta symbol δi,j is definedby declaring δi,i = 1, and δi,j = 0 for i 6= j. More elegantly, this can be rephrased by sayingA ∈ O(n) if and only if AtA = I, where At is the transpose of A, and I is the identitymatrix. In particular, if we define

f : Mn×n −→Mn×n by A 7→ AtA,

then O(n) = f−1(I). Since f is continuous (since each component is), and the one-elementsubset {I} ⊂Mn×n = Rn2

is closed, it follows that O(n) is a closed subset of Rn2.

To show that SO(n) is a closed subset, we note that SO(n) = O(n) ∩ SLn(R), whereSLn(R) is the special linear group consisting of all n × n matrices with determinant 1. Inother words, SLn(R) = det−1(1), where det : Mn×n → R is the map that sends a matrix toits determinant. Since the determinant is a polynomial function of the entries of the matrix,this is a continuous function, and hence SLn(R) = det−1(1) is a closed subset of Rn2

. Thisimplies that the intersection SO(n) = O(n) ∩ SLn(R) is closed as well.

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4. Let X be a topological space which is the union of two subspaces X1 and X2. Let f : X →Y be a (not necessarily continuous) map whose restriction to X1 and X2 is continuous.

(a) Show f is continuous if X1 and X2 are open subsets of X.

(b) Show f is continuous if X1 and X2 are closed subsets of X.

(c) Give an example showing that in general f is not continuous.

Remark. This result is needed to verify that various constructions (e.g., concatenations ofpaths) in fact lead to continuous maps. In a typical situation, we have continuous mapsf1 : X1 → Y and f2 : X2 → Y which agree on X1 ∩X2 and hence there is a well-defined map

f : X −→ Y given by f(x) =

{f1(x) x ∈ X1

f2(x) x ∈ X2

The above result then helps to show that this map is continuous.

Proof. To prove part (a) let U ⊂ Y be open. Then Ui := f−1i (U) is an open subset of Xi,

i.e., it is of the form Ui = Vi ∩Xi, where Vi ⊂ X is open. If follows that Ui is an open subsetof X, and hence f−1(U) = f−1

1 (U) ∪ f−12 (U) = U1 ∪ U2 is an open subset of X, proving the

continuity of f .To prove part (b) we note that f : X → Y is continuous if and only if f−1(U) is a closed

subset of X for every closed subset U ⊂ Y . Then repeating the previous sentences with openreplaced by closed provides a proof of part (b).

For part (c) consider the map f : X = R→ Y = R given by f(t) = 0 if t ∈ (−∞, 0), andf(t) = 1 if t ∈ [0,∞). The restrictions of f to (−∞, 0) resp. [0,∞) are constant and hencecontinuous, but f is not.

5. Which of the topological groups GLn(R), O(n), SO(n) are connected? Hint: Use withoutproof the fact that every element in SO(n) is represented by a matrix of block diagonal formwhose diagonal blocks are the 1× 1 matrix with entry +1 and/or 2× 2 rotational matrices

R =

(cosφ − sinφsinφ cosφ

).

Here “block diagonal” means that all other entries are zero.

Proof. The determinant function det : Mn×n → R is a continuous function. Since the de-terminant for matrices in GLn(R) or O(n) is non-zero, its restriction to G = GLn(R), O(n)provides a continuous map

f : G −→ R \ {0}.

It follows that G = f−1((−∞, 0))∪f−1((0,∞)) is a decomposition of G into a disjoint unionof open subsets. Both of these are not empty, since the determinant of the identity matrixis 1, and the determinant of the diagonal matrix with diagonal entries (−1, 1, . . . , 1) is −1.This shows that the spaces GLn(R) and O(n) are both not connected.

To prove that SO(n) is connected it suffices to show that SO(n) is path connected,i.e., any two elements A,B ∈ SO(n) can be connected by a path. We will show this by

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constructing for every A ∈ SO(n) a path γA : [0, 1] → SO(n) with γA(0) = I (the identitymatrix) and γA(1) = A. This implies that any two points A,B ∈ SO(n) can be connected bya path γAB, obtained by first taking a path from A to I, by running the path γA backwards,and then taking the path γB from I to B. In formulas, the path γAB is given by

γAB(t) =

{γA(1− 2t) for t ∈ [0, 1

2]

γB(2t− 1) for t ∈ [12, 1]

This is in fact a path (i.e., a continuous map) by the previous problem.Let us first construct γA for n = 2, using the fact that every A ∈ SO(2) is a rotation,

given by a matrix of the form

A =

(cos θ − sin θsin θ cos θ

).

We define

γA : [0, 1] −→ SO(2) by γA(t) :=

(cos tθ − sin tθsin tθ cos tθ

)which is continuous since it components are continuous functions of t, and has the desiredproperty γA(0) = I and γA(1) = A.

For a general dimension n, we use the fact that for any A ∈ SO(n) there is a choiceof basis for Rn such that the matrix representing the isometry A is of block diagonal formas described in the hint. Replacing for each block Ri the rotation angle θi (which may bedifferent for the different blocks) by tθi we obtain a path γA : [0, 1] → SO(n) from I to A,generalizing our construction in the n = 2 case.


1. Show that the connected sum RP2#RP2 of two copies of the real projective plane ishomeomorphic to the Klein bottle K.

Proof. Consider the topological spaces given by identifying egdes with the same label for thedisjoint union of polygons given by the following pictures.

b

b

a a ≈ a

b

ca

b

c ≈ a

b

ca

b

c ≈

c

b

b c

The left picture is our standard construction of the Klein bottle K. The second and thirdpicture shows the quotient obtained from two disjoint triangles by identifying edges with thesame label. The second space is homeomorphic to the first since identifying the two edgeslabeled “c” in the second picture, we obtain the first. The third space is homeomorphic tothe second, since the top right triangle of picture 2 is just redrawn as the left triangle of

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picture 3 (flipped over such that the a-edge of that triangle lines up with the a-edge of thesecond triangle). The fourth space is homeomorphic to the third since it is obtained fromthe third space by identifying the two edges labeled “a” in the third picture.

Reading off the word bbcc by starting at the top left corner the last picture and goingaround counterclockwise, the topological space given by that picture is Σ(bbcc). From classwe know the homeomorphisms

Σ(bbcc) ≈ Σ(bb)#Σ(cc) ≈ RP2#RP2.

Composing all the homeomorphisms mentioned above we obtain the desired homeomorphismbetween the Klein bottle K and the connected sum RP2#RP2.

2. a) Show that the surface of genus g

Σg = T# . . .#T︸︷︷︸g

and the connected sumRP2# . . .#RP2︸︷︷︸

k

are both homeomorphic to the topological space Σ(W ) associated to a word W . What isthe word W in these two cases?b) Calculate the Euler characteristic of these manifolds.

Proof. Part (a) From class we know that

• T ≈ Σ(aba−1b−1), RP2 ≈ Σ(cc), and

• Σ(W )#Σ(W ′) if W,W ′ are words from disjoint alphabets.

This impliesΣg = T# . . .#T ≈ Σ(a1b1a

−11 b−1

1 )# . . .#Σ(agbga−1g b−1

g )

≈ Σ(a1b1a−11 b−1

1 . . . agbga−1g b−1

g )(3.1)

andRP2# . . .#RP2 ≈ Σ(a1a1)# . . .#Σ(akak) ≈ Σ(a1a1 . . . akak). (3.2)

Part (b). The homeomorphism 3.1 implies that Σg is the quotient space of a polygon with4g edges (the length of the word a1b1a

−11 b−1

1 . . . agbga−1g b−1

g ) and edge identification encodedby this word. After gluing, this polygon will provide a “pattern of polygons” on Σg withone face and 2g edges (of the original 4g edges of the polygon, the pairs of edges with thesame label are glued, resulting in 2g edges after gluing). Each of the original 4g vertices isindentified with any other by gluing, resulting in one vertex in the quotient space (this iseasy to check for the first four vertices corresponding to the first torus, and then arguinginductively). It follows that

χ(Σg) = #V −#E + #F = 1− 2g + 1 = 2− 2g.

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The argument for RP2# . . .#RP2 is completely analogous. The homeomorphism 3.2 impliesthat this surface is the quotient space of a 2k-gon with edge identifications encoded in theword a1a1 . . . akak. Using the resulting “pattern of polygons” on RP2# . . .#RP2 to calculateits Euler characteristic, we obtain

χ(RP2# . . .#RP2︸︷︷︸k

) = #V −#E + #F = 1− k + 1 = 2− k.

3. Important examples of quotient spaces are orbit spaces of a group G acting on a topologicalspace X. We recall that a (left) of a group G on a set X is given by a map

G×X −→ X typically written as (g, x) 7→ gx,

such that g1(g2x) = (g1g2)x for g1, g2 ∈ G, x ∈ X (associativity) and ex = x for e theidentity element element of G, x ∈ X (identity element property). Given a G-action on atopological space X the orbit space denoted X/G is the quotient space X/ ∼ where twoelements x, y ∈ X are declared equivalent if and only if there is some g ∈ G with gx = y. Inparticular, the equivalence class of x is the subset {gx | g ∈ G}, which is called the orbit ofx, and hence X/G is the space of orbits. Although we haven’t used this terminology, we’vealready encountered orbit spaces, namely

RPn = Sn/{±1} and CPn = S2n+1/S1.

Here the actions are given by

{±1} × Sn → Sn

(t, (v0, . . . , vn)) 7→ (tv0, . . . , tvn)and

S1 × S2n+1 → S2n+1

(z, (z0, . . . , zn)) 7→ (zz0, . . . , zzn),

where z ∈ S1 ⊂ C and (z0, . . . , zn) ∈ S2n+1 ⊂ Cn+1.

(a) Consider the action Z2 × R2 → R2, (m,n), (x, y) 7→ (x + m, y + n). Show that thequotient space R2/Z2 is homeomorphic to the torus, described as the quotient space

b

b

a a

Use without proof the fact that this orbit space is Hausdorff (this will come up later thissemester).

(b) Consider the action G×R2 → R2 where G is the subgroup of the group of isometries ofthe metric space R2 generated by the isometries g, h : R2 → R2 defined by

g(x, y) = (x+ 1, y) and h(x, y) = (−x, y + 1)

11

Show that the quotient space R2/G is homeomorphic to the Klein bottle, described asthe quotient space of the square [0, 1]× [0, 1] with edge identifications

b

b

a a

Again, use without proof the fact that this orbit space is Hausdorff. Hint: Show thatevery orbit can be represented by a point (x, y) ∈ [0, 1]× [0, 1]. To do this, it might behelpful to argue that the composition ghgh−1 is the identity and to use this to show thatevery element of G can be uniquely written in the form gmhn with m,n ∈ Z.

Proof. Part (a). We recall that the torus T is the quotient (I × I)/ ∼ where I = [0, 1] andthe equivalence relation is generated by (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y). Consider the map

f : T = (I × I)/ ∼−→ R2/Z2 [(x, y)] 7→ Z2(x, y),

where we write Z2(x, y) for the orbit of the Z2-action through (x, y).Next we will argue that f is well-defined, surjective, injective and continuous. This

implies that f is a homeomorphism since its domain is compact (as the image of the closedbounded subset I × I ⊂ R2 under the projection map I × I → (I × I)/ ∼, and we use thefact that R2/Z2 is Hausdorff.

• The map f is well-defined since the points (x, 0), (x, 1) ∈ I×I belong to the same orbitof the Z2-action on R2, since (0, 1)(x, 0) = (x, 1). Similarly, (0, y) and (1, y) belong tothe same Z2-orbit.

• For every (x, y) ∈ R2 the element (x− [x], y− [y]) belongs to I×I, where [x] ∈ Z is thelargest integer smaller or equal to x ∈ R. Since ([x], [y]) ∈ Z2 applied to (x−[x], y−[y])gives (x, y), this shows that (x − [x], y − [y]) and (x, y) belong to the same orbit andhence the map f is surjective.

• Two elements (x, y), (x′, y′) ∈ R2 belong to the same Z2-orbit if and only if x− x′ ∈ Zand y − y′ ∈ Z. Hence the only points (x, y) ∈ I × I mapping to the same orbit are ofthe form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I × I/ ∼, themap f is injective.

• To prove continuity of f consider the commutative diagram

I × I i //

p

��

R2

q

��(I × I)/ ∼ f // R2/Z2

.

The inclusion map i and the projection maps p, q are continuous, and so the composi-tion q ◦ i is continuous. By commutativity of the diagram it follows that f ◦ p = q ◦ iis continuous, and hence f is continuous.

12

Part (b). The proof is analogous to that of part (a). Explicitly the Klein bottle K is thequotient (I × I)/ ∼ where the equivalence relation is generated by (x, 0) ∼ (1 − x, 1) and(0, y) ∼ (1, y). Consider the map

f : T = (I × I)/ ∼−→ R2/Z2 [(x, y)] 7→ G(x, y),

where we write G(x, y) for the orbit of the G-action through (x, y). We will prove this inthe same steps as in (a), but first we prove the following result.

Lemma 3.3. The G-orbit G(x, y) of a point (x, y) ∈ R2 is given explicitly by

G(x, y) = {(m+ (−1)nx, n+ y) | m,n ∈ Z}.

To prove this result we first verify that the relation ghgh−1 = 1 holds in G:

(x, y)h−1

7→ (−x,−1 + y)g7→ (1− x,−1 + y)

h7→ (−(1− x), 1 + (−1 + y)) = (−1 + x, y)g7→ (x, y)

A general element of G is a word made from the letters g, g−1, h, h−1. The relation justproved implies hgh−1 = g−1 which implies

h±1g±1 = g∓1h±1.

In particular, we can move all factors g±1 to the left of the factors h±1, which shows thatevery element of G can be written in the form gmhn for m,n ∈ Z. Since

gmhn(x, y) = gm((−1)nx, n+ y) = (m+ (−1)nx, n+ y),

this proves the lemma.

• The map f is well-defined since by the lemma the points (x, 0), (x, 1) ∈ I × I (resp.the points (0, y) and (1, y)) belong to the same G-orbit.

• The map f is surjective since by the lemma for the orbit through any point (x, y) ∈ R2,there is a unique point (x′, y′) ∈ [0, 1)× [0, 1) in the same orbit as (x, y).

For every (x, y) ∈ R2 the element (x− [x], y− [y]) belongs to I×I, where [x] ∈ Z is thelargest integer smaller or equal to x ∈ R. Since ([x], [y]) ∈ Z2 applied to (x−[x], y−[y])gives (x, y), this shows that (x − [x], y − [y]) and (x, y) belong to the same orbit andhence the map f is surjective.

• Two elements (x, y), (x′, y′) ∈ R2 belong to the same Z2-orbit if and only if x− x′ ∈ Zand y − y′ ∈ Z. Hence the only points (x, y) ∈ I × I mapping to the same orbit are ofthe form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I × I/ ∼, themap f is injective.

• To prove continuity of f consider the commutative diagram

I × I i //

p

��

R2

q

��(I × I)/ ∼ f // R2/Z2

.

The inclusion map i and the projection maps p, q are continuous, and so the composi-tion q ◦ i is continuous. By commutativity of the diagram it follows that f ◦ p = q ◦ iis continuous, and hence f is continuous.

13

4. a) Show that a subspace of a Hausdorff space is Hausdorff and that a product of Hausdorffspaces is Hausdorff.b) Show that a subspace of a second countable space is second countable and that a productof second countable spaces is second countable.

Proof. Part (a). Suppose A ⊂ X is a subspace of a Hausdorff space X. To show thatA is Hausdorff, let x, y ∈ A be distinct points. Since X is Hausdorff, there are disjointopen subsets U, V ⊂ X with x ∈ U and y ∈ V . Then U ∩ A and V ∩ A are disjoint openneighborhoods of x resp. y in A, proving that A is Hausdorff.

Suppose X, Y are Hausdorff spaces. To show that X×Y is Hausdorff, let (x, y), (x′, y′) ∈X × Y be distinct points. Without loss of generality we may assume x 6= x′. This allowsus to find disjoint open neighborhoods U, V ⊂ X of x resp. x′. Then U × Y and V × Y aredisjoint open neighborhoods of (x, y) resp. (x′, y′), which proves that X × Y is Hausdorff.

Part (b). Let A be a subspace of a second countable space X. Let {Ui}i∈I be a countablebasis for X. We claim that the countable collection {Ui∩A}i∈I is a basis for A, thus provingthat A is second countable. To prove the claim, we recall that every open subset of A is ofthe form U ∩ A for some open subset U . Now U can be expressed as a union of Ui’s for ibelonging to some subset J ⊂ I. It follows that

U ∩ A = (⋃i∈J

Ui) ∩ A =⋃i∈J

(Ui ∩ A),

proving that {Ui ∩ A}i∈I is a basis for the topology of A.Let X, Y be second countable spaces with countable bases {Ui}i∈I and {Vj}j∈J , respec-

tively. We claim that the countable collection {Ui × Vj}(i,j)∈I×J is a basis for X × Y , thusshowing that X × Y is second countable. To prove the claim, we need to show that everyopen neighborhood W of (x, y) ∈ X×Y contains a neighborhood of (x, y) of the form Ui×Vj.By definition of the product topology, W contains a neighborhood of the form U ×V , whereU is an open neighborhood of x ∈ X and V is an open neighborhood of y ∈ Y . Since theUi’s form a basis for the topology of X, this implies that there is some Ui with x ∈ Ui andUi ⊂ U . Similarly, we find some Vj with y ∈ Vj ⊂ V . Then

(x, y) ∈ Ui × Vj ⊂ U × V ⊂ W

which finishes the proof.


1. Let ωn : (I, ∂I)→ (S1, 1) be the based loop defined by ωn(s) = e2πins. Show that the mapΦ: Z→ π1(S1, 1) given by n 7→ [ωn] is a group homomorphism.

Proof. We note that

Φ(m+ n) = [ωm+n], and Φ(m)Φ(n) = [ωm][ωn] = [ωm ∗ ωn]

14

and so we need to construct a homotopy relative endpoints between ωm+n and the concate-nation ωm ∗ ωn (note that they are not the same path unless m = n: both wind around thecircle m+n times, but ωm+n goes around the circle at a constant speed, while ωm ∗ωn windsaround the circle n times on the interval [0, 1/2], while its winds around m times on [1/2, 1]).To produce the desired homotopy we note that ωm+n = p ◦ α1 and ωm ∗ ωn = p ◦ α2, wherep : R→ S1 is defined by p(s) = e2πis, and the paths αi : [0, 1]→ R are given by

α1(s) = (m+ n)s α2(s) =

{2ns 0 ≤ s ≤ 1/2

2ms−m+ n 1/2 ≤ s ≤ 1

We note that paths α1 and α2 have the same start/end point and hence they are homotopicvia the linear homotopy Ht connecting them. Then p ◦Ht is the desired homotopy betweenthe paths ωm+n = p ◦ α1 and ωm ∗ ωn = p ◦ α2.

2. Let γ : I → X be a path in a topological space X, and let γ : I → X be the path γ runbackwards, that is, γ(s) = γ(1− s). Then there are the following homotopies

γ ∗ γ ' cγ(0) γ ∗ γ ' cγ(1) γ ∗ cγ(0) ' γ cγ(1) ∗ γ ' γ, (1)

where cx for x ∈ X denotes the constant path at x.

(a) Prove the first and the third homotopy.

(b) Use the homotopies (1) to finish our proof that π1(X, x0) is a group. In other words,show that the constant map cx0 gives an identity element for π1(X, x0), and that [γ] isthe inverse for [γ] ∈ π1(X, x0).

(c) Let β be a path from x0 to x1. Show that the map

Φβ : π1(X, x0) −→ π1(X, x1) [γ] 7→ [β ∗ γ ∗ β]

is an isomorphism of groups.

Proof. Part (a). The strategy of constructing these homotopies is analoguous to the strategyused in problem 1. We note that

γ ∗ γ = γ ◦ α1 and cγ(0) = γ ◦ α2,

where α1, α2 : [0, 1]→ [0, 1] are the paths in [0, 1] given by

α1(s) =

{2s 0 ≤ s ≤ 1

2

2(1− s) 12≤ s ≤ 1

and α2(s) = 0.

Let Ht be the linear homotopy between α1 and α2. This is a homotopy of paths since theinitial and terminal points of the paths α1 and α2 agree. This implies that γ ◦ Ht is thedesired homotopy between γ ◦ α1 = γ ∗ γ and γ ◦ α1 = cγ(0).

Similarly,γ ∗ cγ(0) = γ ◦ β1 and γ ∗ cγ(0) = γ ◦ β2,

15

where β1, β2 : [0, 1]→ [0, 1] are the paths given by

β1(s) =

{0 0 ≤ s ≤ 1

2

2(s− 1) 12≤ s ≤ 1

and β2(s) = 0. As before, the linear homotopy Gt between these maps is a path homotopysince β1(0) = β2(0) and β1(1) = β2(1). It follows that γ ◦Gt is a homotopy between γ ∗ cγ(0)

and γ as desired.

Part (b). Let γ be a loop in X, based at x0 ∈ X. Then the third and fourth homotopy inpart (a) implies the equalities in π1(X, x0)

[γ][cx0 ] = [γ] [cx0 ][γ] = [γ].

This shows that [cx0 ] is the identity element for the multiplication in π1(X, x0).Similarly, the first and second homotopy from part (a) imply the equalities

[γ][γ] = [cx0 ] [γ][γ] = [cx0 ]

in π1(X, x0). These equalities imply that [γ] is the inverse of [γ]. In particular, π1(X, x0) isa group.

Part (c). First we show that Φβ : π1(X, x0) −→ π1(X, x1) is a homomorphism. So let γ, δbe loops in X based at x0. Then

Φβ([γ][δ]) = Φβ([γ ∗ δ]) = [β ∗ γ ∗ δ ∗ β] = [β ∗ γ ∗ β ∗ β ∗ δ ∗ β]

= [β ∗ γ ∗ β][β ∗ δ ∗ β] = Φβ([γ])Φβ([δ])

Here we avoid using parantheses for the concatenation operation; this is OK by the factthat the paths resulting from different ways of putting in parantheses leads to homotopicpaths. The first and second equality is just the definition of multiplication in π1(X, x0)resp. the definition of Φβ. The third equation holds due to the homotopy β ∗ β ' cx0 (anapplication of the second homotopy of part (a)), and the fact that we can insert at any pointan appropriate constant path into an iterated concatenation without changing the homotopyclass of the path (follows from either the third or the fourth homotopy of part (a)). The lasttwo equations again hold by construction of multiplication in the fundamental group resp.the map Φβ.

It remains to show that Φβ is an isomorphism. We claim that its inverse is given byΦβ : π1(X, x1)→ π1(X, x0).

ΦβΦβ(γ) = Φβ([β ∗ γ ∗ β]) = [β ∗ β ∗ γ ∗ β ∗ β] = [cx1 ∗ γ ∗ cx0 ] = [γ]

Here the first two equations is just by construction of Φβ resp. Φβ, the third equation holdsthanks to the first two homotopies of part (a), and the last equation holds due to the lasttwo homotopies of part (a).

3. A subspace A ⊂ X of a topological space X is called a retract of X if there is a continuousmap r : X → A whose restriction to A is the identity.

16

(a) Show that S1 is not a retract of D2. Hint: Show that the assumption that there is acontinuous map r : D2 → S1 which restricts to the identity on S1 leads to a contradictionby contemplating the induced map r∗ of fundamental groups.

(b) Brouwer’s Fixed Point Theorem states that every continuous map f : Dn → Dn has afixed point, i.e., a point x with f(x) = x. Prove this for n = 2. Hint: show that if f hasno fixed point, then a retraction map r : D2 → S1 can be constructed out of f .

Proof. We prove part (a) be contradiction. Suppose that r : D2 → S1 is a retraction of D2

onto S1. This means that r makes the following diagram commutative:

S1 i //

id !!BBB

BBBB

B D2

r��S1

Choosing a basepoint x0 ∈ S1 and applying the fundamental group functor gives the com-mutative diagram

π1(S1, x0)i∗ //

id∗=id ''NNNNN

NNNNNN

π1(D2, x0)

r∗��

π1(S1, x0)

This gives the desired contradiction since according to the diagram the identity map onπ1(S1, x0) ∼= Z factors through the trivial group π1(D2, x0).

To prove part (b) we assume that f : Dn → Dn does not have a fixed point, i.e., that forall x ∈ Dn, f(x) 6= x. Then the line through the points f(x) and x intersects the sphereSn−1 in two points. Let r(x) ∈ Sn−1 be the intersection point closer to x. It is clear from theconstruction that r : Dn → Sn−1 has the property r(x) = x for x ∈ Sn−1, and so it remainsto show the continuity of r to derive the desired contradiction from part (a).

To show that r is continuous, we note that r(x) can be written in the form

r(x) = x+ α(x)(x− f(x)),

where α(x) ∈ R is the unique non-negative solution of the quadratic equation

||x+ α(x)(x− f(x))||2 = 1 (4.1)

(the solutions of this equation correspond to the two intersections of the line through xand f(x) with Sn−1; hence it is clear geometrically, that there are two solution for everyx, and that exactly one solution is non-negative). Writing out the quadratic equation (4.1)explicitly as

||x− f(x)||2α2(x) + 2〈x, x− f(x)〉α(x) + ||x||2 − 1 = 0

shows that its coefficients are continuous functions of x, and hence the quadratic formulashows that its non-negative solution α(x) is continuous function of x. We conclude that r(x)is a continuous function of x since its components are linear combinations of products ofcontinuos functions.

17

4. We recall that the torus T and the Klein bottle K can both be described as quotientspaces of the square by identifying edges with the same label as shown in the followingpictures.

T =

b

b

a a K =

b

b

a a

Each edge of the square can be interpreted as a path in the square, which projects to a basedloop in the quotient space T resp. K. Here the base point x0 of the quotient space is theprojection of each of the four vertices of the square. Abusing language, let us denote by aresp. b the elements in the fundamental group of T (resp. K) represented by these loops.Prove the following identities:

aba−1b−1 = 1 ∈ π1(T, x0) abab−1 = 1 ∈ π1(K, x0).

Proof. We begin with the torus T = ([0, 1] × [0, 1])/ ∼ with equivalence relation generatedby (0, y) ∼ (1, y) and (x, 0) ∼ (x, 1). For i = 1, 2, let αi, βi : I → [0, 1] × [0, 1] be the pathsin the square [0, 1] × [0, 1] given by its edges, with the direction provided by the arrow oneach edge, as shown in the picture

[0, 1]× [0, 1] =

β1

β2

α1 α2

x0

It should be emphasized that this is a precise description by adopting the convention thatall of these paths are the linear paths connecting their endpoints. Going counterclockwisearound the boundary of the square, we obtain the loop

γ := α1 ∗ β1 ∗ α2 ∗ β2,

based at the top left corner point x0 (strictly speaking we should put parentheses here, sincewe are talking about paths, but we don’t want to burden the notation). Let p : I×I → T bethe projection map. We note that the projection p ◦ αi of the paths αi is a loop in T basedat the point x0 := p(x0). In fact, p ◦ α1 and p ◦ α2 are the same loop in T which we denoteby α, and [α] ∈ π1(T, x0) is the generator called “a”. Similarly, projecting the paths βi weobtain a loop β in T based at x0 with [β] = b ∈ π1(T, x0). Hence

aba−1b−1 = [α][β][α][β] = [α ∗ β ∗ α ∗ β] = [p ◦ (α1 ∗ β1 ∗ α2 ∗ β2)] = [p ◦ γ].

Since γ is a loop based at x0 in the convex subset [0, 1]2 ⊂ R2, the linear homotopy Ht

between this concatenation and the constant loop cx0 is a homotopy relative endpoints (we

18

want to emphasize that the fact that γ is a loop is essential here; by contrast, the linearhomotopy between the path α1 and cx0 is not a homotopy relative endpoints). It followsthat p ◦ Ht : I → T is a homotopy between p ◦ γ and the constant loop cx0 , which by (??)implies the relation aba−1b−1 = 1 in the fundamental group of the torus.

The argument in the case of the Klein bottle K is completely analogous. As before wehave paths αi, βi for i = 1, 2 in [0, 1]2 which project to loops α, β based at x0 = p(x0) ∈ Kunder the projection map p : [0, 1]2 → ([0, 1]2/ ∼) = K. Here is the picture:

[0, 1]× [0, 1] =

β1

β2

α1 α2

x0

The only difference to the previous situation is that we have the loop

δ := α1 ∗ β1 ∗ α2 ∗ β2

in [0, 1]2, which is homotopic to the constant loop cx0 . The image of the loop δ under p isthe loop α ∗ β ∗ α ∗ β which represents the element abab−1 in the fundemental group of theKlein bottle. This proves the relation abab−1 = 1.

5. Let Z2 act on R2 by translation (as in problem # 3 of the last set). Show that theprojection p : R2 → R2/Z2 is a covering map.

Proof. We note that for (x, y) ∈ R2 the subset p−1(p(x, y)) ⊂ R2 is the orbit through (x, y)(this is true in general for the projection map p : X → X/G to the orbit space for a G-actionon a topological space X). More explicitly:

p−1(p(x, y)) = {(x+m, y + n) ∈ R2 | m,n ∈ Z}.

In particular, if U ⊂ R2 is an open subset, then

p−1(p(U)) =⋃

(m,n)∈Z2

(m+ n) + U (4.2)

is an open subset of R2, since it is the union of the open subsets (m,n) + U given by theimage of U under the homeomorphism R2 → R2, (x, y) 7→ (x + m, y + n). It follows thatp(U) is an open subset of R2/Z2. In other words, p is an open map, which means that ptakes open subsets of its domain to open subsets of its range.

We claim that the open subset p(U) ⊂ R2/Z2 is evenly covered, provided the intersection((m,n)+U)∩U is empty for (m,n) 6= (0, 0). To prove the claim, assume that ((m,n)+U)∩U = ∅ for (m,n) 6= (0, 0). This implies that the the sets (m,n) + U in the decomposition(4.2) of p−1(p(U)) are pairwise disjoint, since if (x, y) ∈ ((m,n) + U) ∩ ((m′, n′) + U) for(m,n) 6= (m′, n′), then (x − m′, y − n′) ∈ ((m − m′, n − n′) + U) ∩ U , contradicting ourassumption that this intersection is empty. This implies that the restriction of p to (m,n)+U

19

is a continuous bijection between (m,n) + U and p(U). Moreover, this map is open (as therestriction of the open map p), and hence it is a homeomorphism. This finishes the proof ofthe claim above.

To finish the proof we note that for any (x, y) ∈ R2 the open ball U = B1/2(x, y) of radius1/2 around (x, y) does not intersect any translated ball (m,n) +U . In particular p(U) is anevenly covered open subset. Since every point of R2/Z2 is contained in an open subset ofthis form, this proves that p : R2 → R2/Z2 is a covering map.


1. Let f : S1 → S1 be defined by f(z) = zn for some n ∈ Z. Calculate the inducedhomomorphism

f∗ : π1(S1, 1) −→ π1(S1, 1).

Clarification: We’ve proved that the fundamental group π1(S1, 1) is isomorphic to Z. Inparticular, any endomorphism of π1(S1, 1) is given by multiplication by some integer k ∈ Z.“Calculating f∗” means determining that integer k for the endomorphism f∗ ∈ End(π1(S1, 1)).

Proof. We recall that the homotopy class [ω1] ∈ π1(S1, 1) of the based loop ω1 : I → S1,defined by ω1(s) = e2πis, is a generator of π1(S1, 1). So it suffices to calculate f∗[ω1] =[f ◦ ω1] ∈ π1(S1, 1):

f ◦ ω1(s) = f(e2πis) = (e2πis)n = (e2πins) = ωn(s).

Hence f∗([ω1]) = [ωn] = n[ω1] ∈ π1(S1, 1), where the second equation was proved in problem#1 of assignment # 4. This shows that the endomorphism f∗ is given by multiplication byn ∈ Z.

2. Let p : X −→ X be a covering space such that for each x ∈ X the fiber p−1(x) is countable.

(a) Show that if X is a manifold of dimension n, then so is X. Hint: to show that X hasa countable basis, first show that X has a countable basis consisting of evenly coveredopen subsets.

(b) If X is compact manifold of dimension 2 and p is a d-fold covering map, what is the

Euler characteristic of X in terms of the Euler characteristic of X? Here a d-fold coveringmeans that each fiber p−1(x) consists of d ∈ N points.

(c) Suppose that p : X −→ X is a d-fold covering, d ∈ N where X is a surface of genus g

and X is a surface of genus g. Give a formula expressing g in terms of g and d.

Proof. Part(a). Given a point x ∈ X, let x = p(x) ∈ X, and let U ⊂ X be an open

neighborhood of x and φ : U≈−→ V be a homeomorphisms between U and an open subset

V ⊂ Rn. Replacing U by a smaller neighborhood of x if necessary, we may assume that U is

20

evenly covered. Then p−1(U) is the disjoint union of open subsets Uα such that p|Uα : Uα → Uis a homeomorphism. Then x is contained in some Uα, and the composition

Uαp|Uα // U

φ // V ⊂ Rn

provides a homeomorphism between Uα and V . This proves that X is locally homeomorphicto Rn.

To show that X is Hausdorff, let x, y ∈ X with x 6= y. If p(x) 6= p(y), then we can finddisjoint open neighborhoods Ux, Uy ⊂ X of x := p(x) resp. y := p(y), and then p−1(Ux),p−1(Uy) are disjoint open neighborhoods of x resp. y. If x = y, let U be an evenly coveredneighborhood of x, and let p−1(U) =

∐α∈A Uα as above. If x belongs to Uα, then y belongs

to Uβ for β 6= α, and hence Uα, Uβ are disjoint open neighborhoods of x and y.

To show that X is second countable, let B′ be a countable basis for the topology of X,and let B be the subcollection of those open sets U ⊂ X belonging to B′ which are evenlycovered. We claim that this is a basis for the topology on X. To prove this, let x ∈ X andlet V be an open neighborhood of x. Intersecting with an evenly covered open set containingx, we may assume that V is evenly covered. Since B′ is a basis, there is some U ∈ B′ withx ∈ U and U ⊆ V . Since V is evenly covered, so is U , and hence U belongs to B. Thisproves that B is a basis for the topology on X; as a subcollection of the countable collectionB′ it is countable.

Now we construct a countable basis B of X as follows. For each U ∈ B, pick a pointxU ∈ U . By construction U is evenly covered, and hence p−1(U) is a disjoint union of open

subsets of X, such that the restriction of p to each of these subsets is a homeomorphism. Inparticular, each of these subsets contains exactly one of the points x ∈ p−1(xU), allowing us

to index these sets by the elements of p−1(xU): we will write Ux for the subset containingthe point x ∈ p−1(xU). With this notation, we have

p−1(U) =∐

x∈p−1(xU )

Ux,

and p|Ux : Ux −→ U is a homeomorphism. Let B be the following collection of subsets of X:

B := {Ux | U ∈ B, x ∈ p−1(xU)}.

This is a countable collection, since B is countable, and each fiber p−1(xU) is countable.

We claim that B is a basis for X. To show this, it suffices to prove that for every y ∈ Xand open neighborhood V ⊆ X there is some Ux ∈ B with y ∈ Ux and Ux ⊆ V . Lety := p(y), let U ∈ B an open set containing y, and let Ux ∈ B be the subset of p−1(U) thatcontains the point y. The open neighborhood Ux of y might not yet be what we are lookingfor; it might be too big in the sense that it might not be a subset of V . In order to fix this,consider Ux ∩ V , a smaller open neighborhood of y that via p maps homeomorphically to anopen neighborhood of y. Then there is smaller open neighborhood U ′ of y belonging to thebasis B, and hence for some x′ ∈ p−1(xU

′) the open subset Ux′ is an open subset belonging

to B, which contains the point y and is contained in V . This shows that B is indeed a basisfor the topology of X.

21

Part (b). Let P be a pattern of polygons on X. By passing to a refinement of P is necessarywe can assume that each face is contained in an evenly covered open subset of X. Then thepreimages of vertices, edges and faces of P will form a pattern of polygons P on X. Sincethe preimage of each vertex (resp. edge resp. face) of P consists of d vertices (resp. edges

resp. faces) of P . In particular, if we denote by V (resp. E resp. F ) the number of vertices

(resp. edges resp. faces) of P and by V (resp. E resp. F ) the corresponding number for P ,we have

χ(X) = V − E + F = dV − dE + dF = dχ(Σ).

Part (c). We recall that the Euler characteristic of a surface Σg of genus g is given by

χ(Σg) = 2− 2g. This implies

g =1

2(2− χ(Σg)).

It follows that

g =1

2(2− χ(X)) =

1

2(2− dχ(X)) =

1

2(2− d(2− 2g)) = g − d+ 1.

3. Show that assigning to a pointed space (X, x0) its fundamental group π1(X, x0), andto each basepoint preserving map f : (X, x0) → (Y, y0) the induced map f∗ : π1(X, x0) →π1(Y, y0) defines a functor

π1 : Top∗ −→ Gp

from the category Top∗ of pointed topological spaces to the category Gp of groups.

Proof. To prove that π∗ is a functor, we need to show that it is compatible with compositionsand that it maps identity morphisms in Top∗ to identity morphisms of Gp. To show that itis compatible with compositions, suppose

f : (X, x0)→ (Y, y0) and g : (Y, y0)→ (Z, z0)

are morphisms in Top∗. We need to show that

(g ◦ f)∗ = g∗ ◦ f∗

as group homomorphisms from π1(X, x0) to π1(Z, z0). So let γ : I → X be a based loop in(X, x0) representing the element [γ] ∈ π1(X, x0). Then

(g∗ ◦ f∗)([γ]) = g∗(f∗([γ])) = g∗([f ◦ γ]) = [g ◦ (f ◦ γ)]

(g ◦ f)∗([γ]) = [(g ◦ f) ◦ γ].

The associativity of maps implies that the based loops g ◦ (f ◦ γ) and (g ◦ f) ◦ γ in (Z, z0)agree and proves compatibility with composition.

To show that identity morphisms are preserved, let idX be the identity morphism of thepointed space (X, x0). Then

(idX)∗([γ] = [idX ◦γ] = [γ],

which shows that the induced homomorphism (idX)∗ : π1(X, x0)→ π1(X, x0) is the identityas required.

22

4. Given a set S, let F [S] be the free group generated by S. The elements of F [S] are equiva-lence classes of words whose letters consist of the symbols s and s−1 for s ∈ S (including theempty word). If a word contains letters s and s−1 adjacent to each other, the word obtainedby deleting those two letters is declared equivalent to the original word, and the equivalencerelation is generated by these relations. The multiplication is given by concatenation ofwords, and the identity element is provided by the empty word.

(a) Show that there is a functor F : Set → Gp which on objects sends a set S to the freegroup F [S].

(b) Show that for any group G the map

Gp(F [S], G) −→ Set(S,G) given by f 7→ f|S

is a bijection. Here we identify S with a subset of F [S] by regarding s ∈ S as a oneletter word.

Proof. Part (a). We need to construct the functor F on morphisms, and show that ourconstruction is compatible with composition and identity morphisms.

To construct F on morphisms, we need to associate to each map of sets

f : S → T a group homomorphism F (f) : F (S)→ F (T ).

Every element of F (S) is given by a word sε11 . . . sεkk with si ∈ S and εi ∈ {±1}. We define

F (f) byF (f)(sε11 . . . s

εkk ) := f(s1)ε1 . . . f(sk)

εk . (5.1)

We note that this is well defined. If si+1 = si and εi+1 = −εi, the word sε11 . . . sεkk is equivalent

to the shorter word obtained by removing the segment sεii sεi+1

i+1 from it. Their images underF (f) are again equivalent, since the shorter one is obtained by removing the segment

f(si)εif(si+1)εi+1

from the longer word.Next we need to show that our map F (f) : F (S) → F (T ) is indeed a group homomor-

phism. It is clear that it maps the identity element of F (S), given by the empty word, tothe identity element of F (T ). It is also clear that it maps a product a · b in F (S) to theproduct F (f)(a) · F (f)(b), since the product is given by concatenation of words.

The formula (5.1) shows that if f : S → S is the identity map of S, then F (f) is theidentity map of F (S).

Finally we need to show that F is compatible with compositions. So let f : S → T andg : T → U be maps between sets. Then

(F (g) ◦ F (f))(sε11 . . . sεkk ) = F (g)(F (f)(sε11 . . . s

εkk ))

= F (g)(f(s1)ε1 . . . f(sk)εk)

= g(f(s1))ε1 . . . g(f(sk))εk

= ((g ◦ f)(s1))ε1 . . . ((g ◦ f)(sk))εk

= F (g ◦ f)(sε11 . . . sεkk )

23

Part (b). To show that the map Φ: Gp(F [S], G) −→ Set(S,G) is a bijection, we considerthe map

Ψ: Set(S,G) −→ Gp(F [S], G)

that sends a map g : S → G to the group homomorphism Ψ(g) : F [S]→ G defined by

Ψ(g)(sε11 . . . sεkk ) := g(s1)ε1 . . . g(sk)

εk .

By the same arguments as in part (a) this map is well-defined, and a group homomorphism.Applying the homomorphism Ψ(g) : F [S]→ G to one the one-letter word s ∈ S ⊂ F [S],

we have Ψ(g)(s) = g(s), which shows Φ ◦Ψ = id. Conversely, given a group homomorphismf : F [S]→ G, then (Ψ◦Φ)(f) is another group homomorphism F [S]→ G which agrees withf on the generators s ∈ S ⊂ F [S] of the group F [S], and hence (Ψ ◦ Φ)(f) = f .

5. Let Vα, α ∈ A be a collection of vector spaces over a fixed field k. Let

∏α∈A

Vα :=

{∑α∈A

vαα | vα ∈ Vα

}

be the set of all formal linear combinations of elements in A such that the coefficient vα ofα ∈ A belongs to Vα. This is a k-vector space with addition (resp. scalar multiplication)given by adding coefficients (resp. multiplying all coefficients by a fixed element of k).

(a) Show that∏

α∈A Vα is a product of the objects Vα, α ∈ A in the category Vectk.

(b) Let ⊕α∈A

Vα ⊂∏α∈A

Vα

be the subspace consisting of those linear combinations∑

α∈A vαα such that only finitelymany coefficients vα are non-zero. Show that

⊕α∈A Vα is a coproduct of the objects Vα

in Vectk.

Proof. Part (a). To show that∏

α∈A Vα is a product of the Vα’s, define linear map

pβ :∏α∈A

Vα −→ Vβ by∑α∈A

vαα 7→ vβ.

We need to show that for any vector space W the map

Vect(W,∏α∈A

Vα) −→∏α∈A

Vect(W,Vα) f 7→ (pα ◦ f)α∈A (5.2)

is a bijection. Let us write fα := pα ◦ f ∈ Vect(W,Vα). We note that for w ∈ W we have

f(w) =∑α∈A

fα(w)α ∈∏α∈A

Vα. (5.3)

In particular, f(w) is determined by the collection fα(w) ∈ Vα, which shows that the map(5.2) is injective. Conversely, given a collection of linear maps fα ∈ Vect(W,Vα), the formula

24

(5.3) defines a homomorphism f : Vect(W,∏

α∈A Vα), thus showing that the map (5.2) issurjective.

Part (b). To show that⊕

α∈A Vα is a coproduct of the Vα’s, define for β ∈ A the linearmap

iβ : Vβ →⊕α∈A

Vα by iβ(v) :=∑α∈A

vαα, vα =

{v α = β

0 α 6= β

We claim that for any vector space W the map

Φ: Vect(⊕α∈A

Vα,W ) −→∏α∈A

Vect(Vα,W ) f 7→ (f ◦ iα)α∈A

is a bijection, which implies that⊕

α∈A Vα is a coproduct of the Vα’s. To prove this claimwe consider the map

Ψ:∏α∈A

Vect(Vα,W ) −→ Vect(⊕α∈A

Vα,W )

defined by

(fα)α∈A 7→ (∑α∈A

vαα 7→∑α∈A

fα(vα)).

We note that the sum∑

α∈A fα(vα) gives a well defined vector in W since it is a finite sum,since all but finitely many of the vα’s vanish. It is straightforward to check that the formulaabove defines a linear map from

⊕Vα to W , and that Ψ is the inverse of Φ.


1. Show that the following five topological spaces are all homotopy equivalent:

1. the circle S1,

2. the open cylinder S1 × R,

3. the annulus A = {(x, y) | 1 ≤ x2 + y2 ≤ 2},

4. the solid torus S1 ×D2,

5. the Mobius strip

Hint: show that each of the spaces (2)-(5) contains a subspace S homeomorphic to the circleS1 which is a deformation retract of the bigger space it is contained in.

Proof. As suggested by the hint, for each of the spaces X in cases (2)-(5) we construct acontinuous injection i : S1 → X. This gives a continuous bijection from S1 to the subspaceS := i(S1) ⊂ X. Since S1 is compact and the spaces X under consideration are all Hausdorff,

25

the subspace S is homeomorphic to S1. To show that S is a deformation retract of X weneed to construct a continuous map

H : X × I −→ X

with the properties

(a) H(x, 1) = x for all x ∈ X;

(b) H(x, 0) ∈ S for all x ∈ X;

(c) H(x, t) = x for x ∈ S.

This map is the homotopy between the identity of X and the retraction r : X → S ⊂ Xgiven by r(x) := H(x, 0), showing that S is a deformation retract.

Here are the maps i : S1 → X and H : X × I → X for each of the cases (2)-(5). Itis straightforward to show that the maps i, H are continuous, since we describe them byexplicit formulas, to prove the injectivity of i, and to show that H has properties (a), (b)and (c).

For the cylinder S1 × R we define

i : S1 −→ X = S1 × R H : S1 × R× I −→ S1 × Rz 7→ (z, 0) (z, s, t) 7→ (z, st)

For the annulus A = {(x, y) ∈ R2 | 1 ≤ x2 + y2 ≤ 2} = {z ∈ C | 1 ≤ |z|2 ≤ 2} we define

i : S1 −→ A H : A× I −→ A

z 7→ z (z, t) 7→ (1− t) z|z|

+ tz

For the solid torus S1 ×D2 we define

i : S1 −→ S1 ×D2 H : S1 ×D2 × I −→ S1 ×D2

z 7→ (z, 0) (z, w, t) 7→ (z, tw)

For the Mobius strip M = (I × [−1, 1])/ ∼ with the equivalence relation (0, s) ∼ (1,−s)it will be convenient to replace S1 by the homeomorphic space R/Z. We define

i : R/Z −→M H : M × I −→M

[r] 7→ [r, 0] ([r, s], t) 7→ [r, ts]

2. Let H, G1, G2 be groups, and f1 : H → G1, f2 : H → G2 be homomorphisms. Show that

the amalgamated product G1∗HG2 is the pushout of G1 Hf1oo f2 // G2 , that is, show that

there are group homomorphisms gi : Gi → G1 ∗H G2 such that the commutative diagram

H

f1��

f2 // G2

g2��

G1g1 // G1 ∗H G2

(6.1)

is a pushout diagram.

26

Proof. Let us recall the definition of G1 ∗G2 and G1 ∗HG2. The elements of the free productG1 ∗G2 are equivalence classes of words a1 . . . am whose letters ai are elements of G1 or G2.The amalgamented product G1 ∗H G2 is the quotient of G1 ∗G2 by the normal subgroup Ngenerated by the words of the form f1(h)(f2(h))−1.

Let Gi → G1 ∗G2 for i = 1, 2 be the homomorphism that sends an element g ∈ Gi to theelement of G1 ∗G2 given by the one letter word g ∈ G1 ∗G2, and let gi : Gi → G1 ∗H G2 bethe composition of that homomorphism and the projection map G1 ∗G2 → G1 ∗H G2. Thismakes the diagram (6.1) commutative, since g1 ◦f1 sends an element h to [f1(h)] ∈ G1 ∗H G2

(the element of the quotient G1 ∗HG2 represented by the one letter word f1(h)), while g2 ◦f2

sends an element h to [f2(h)]. These are different elements in G1 ∗ G2, but the quotientG1 ∗H G2 is taylor made for these elements to agree in.

To show that the diagram (6.1) is a pushout diagram we need to check the universalproperty of the pushout, that is, we need to show that for any group K the map

Φ: Gp(G1 ∗H G2, K) −→ Gp(G1)×Gp(H,K) Gp(G2, K) k 7→ (k ◦ g1, k ◦ g2)

is a bijection. To show this, we define a map

Ψ: Gp(G1)×Gp(H,K) Gp(G2, K) −→ Gp(G1 ∗H G2, K)

that is an inverse to Φ. Given (k1, k2) ∈ Gp(G1) ×Gp(H,K) Gp(G2, K), we define Ψ(k1, k2) ∈Gp(G1 ∗H G2, K) by

Ψ(k1, k2)([a1 . . . am]) := k(a1) . . . k(am) where k(a) :=

{k1(a) if a ∈ G1

k2(a) if a ∈ G2

.

It is clear that Ψ(k1, k2) constructed as above is a well-defined homomorphism from G1 ∗G2

to K, but we need to check that Ψ(k1, k2) sends the generators f1(h)f2(h−1) ∈ G1 ∗ G2 ofthe normal subgroup N to the identity element of K:

Ψ(k1, k2)(f1(h)f2(h−1)) = k(f1(h))k(f2(h−1)) = k1(f1(h))k2(f2(h−1))

= k2(f2(h))k2(f2(h−1)) = 1 ∈ K.

Here the third equality holds since (k1, k2) is an element of

Gp(G1)×Gp(H,K) Gp(G2, K) = {(k1, k2) ∈ Gp(G1)× Gp(G2, K) | k1 ◦ f1 = k2 ◦ f2}.

Checking that Ψ is an inverse to Φ is straightforward from the construction of Ψ.

3. Use the Van Kampen Theorem to show π1(Sn, x0) = {1} for n > 1.

Proof. Let U, V be the open subsets of Sn obtained by removing the northpoleN = (0, . . . , 0, 1)resp. the south pole S = (0, . . . , 0,−1). Via the stereographic projection U and V are home-omorphic to Rn, and hence their fundamental group is trivial. We claim that U ∩ V is pathconnected, provided n > 1. This allows us to apply the Van Kampen Theorem which implies

π1(Sn) ∼= π1(U) ∗π1(U∩V ) π1(V ) = {1} ∗π1(U∩V ) {1} = {1}.

27

To prove that U ∩ V = Sn \ {N,S} is pathconnected, suppose that x, y ∈ U ∩ V . We tryto construct a path γ connecting x and y by taking the linear path δ : I → Rn+1 between xand y, given by δ(s) = (1− s)x+ sy, and then dividing by the norm to obtain a path in Sn.In other words,

γ : I → U ∩ V is given by γ(s) =δ(s)

|δ(s)|.

We note that there are two potential issues with this construction: the denominator couldbecome 0 for some s ∈ I and γ(s) could be equal to N or S for some s ∈ I. The firstproblem doesn’t occur if δ doesn’t pass through 0 ∈ Rn+1, and the second problem doesn’toccur if δ(s) doesn’t belong to the subspace spanned by N for any s ∈ I. In particular, if ydoesn’t belong to the 2-dimensional subspace of Rn+1 spanned by x and N , then the curveγ constructed above is a path in U ∩ V connecting x and y.

If n > 1 and x, y ∈ U ∩ V we can choose some z ∈ Sn such that z neither belongs tothe plane spanned by x and N nor the plane spanned by y and N . Then the constructionabove gives paths connecting x and z resp. y and z, and hence their concatenation gives apath between x and y.

4. We recall that if G ×X → X is the action of a group G on a set X, then the subgroupGx := {g ∈ G | gx = x} ⊆ G is the isotropy subgroup of the point x ∈ X. The action iscalled free if the isotropy subgroup Gx is the trivial group for all x ∈ X.

(a) Show that if G is a finite group which acts freely and continuously on a Hausdorff spaceX, then the projection map p : X → X/G to the orbit space X/G is a covering map.Hint: Use the assumptions that the action is free and X is Hausdorff to show that forevery x ∈ X there is an open neighborhood U such that the subsets gU ⊂ X for g ∈ Gare mutually disjoint.

(b) Show that if X is a manifold of dimension n, then also the orbit space X/G is a manifoldof dimension n (don’t forget to argue that X/G is Hausdorff and second countable).

(c) Show that the map

Z/k × S2n−1 −→ S2n−1 given by (m, v) 7→ e2πim/kv

is a continuous free action of the cyclic group Z/k on the sphere S2n−1 ⊂ Cn. By part(b) the orbit space S2n−1/Z/k is then a manifold of dimension 2n− 1, known as a lensspace.

Proof. Part (a). For any point x ∈ X the points gx for g ∈ G are all distinct, since the actionis free (if gx = hx for g 6= h, then h−1gx = x and hence h−1g ∈ Gx, contradicting the freenessassumption). This implies that there are open neighborhoods Ugx of gx ∈ X which aremutually disjoint, since X is Hausdorff. We note that g−1Ugx is another open neighborhoodof x (it is open, since it is the image of the open set Ugx under the homeomorphism givenby the action of the group element g−1). Hence the finite intersection

U :=⋂g∈G

g−1Ugx

28

is again an open neighborhood of x.We claim that the subsets gU for g ∈ G are mutually disjoint. To prove this, assume

that y ∈ gU ∩ hU for g 6= h. Then in particular y belongs to g(g−1Ugx) = Ugx and toh(h−1Uhx) = Uhx contradicting Ugx ∩ Uhx = ∅.

Given the point [x] ∈ X/G the subset V = p(U) ⊂ X/G is an open neighborhood of [x];it is open since p−1(V ) =

⋃g∈G gU is open. This open set is evenly covered, since for any

g ∈ G the restrictionp|gU : gU −→ V

is a homeomorphism, since it is a continuous bijection and an open map. To argue that pis open, suppose that U ⊂ X is an open subset. Then p(U) is an open subset of X/G sincep−1(p(U)) =

⋃g∈G gU is an open subset of X.

Part (b). To show that X/G is locally homeomorphic to Rn, let y ∈ X/G and let V ⊂ X/Gbe an evenly covered open neighborhood of y. Then p−1(V ) is the disjoint of open subsetsVα ⊂ X such that p|Vα : Vα → V is a homeomorphism. Let x be the unique point in Vαwith p(x) = y. Since X is a manifold of dimension n a possibly smaller neighborhood ofx is homeomorphic to an open subset of Rn. Via the restriction of p|Vα to this smallerneighborhood we conclude that a neighborhood of y is homeomorphic to an open subset ofRn.

To show that X/G is second countable, let B be a basis for X. We claim that thecountable collection of subsets of X/G given by p(U) for U ∈ B is a basis for X/G. To seethis, let y ∈ X/G, and let V ⊂ X/G be an open neighborhood of y. Since B is a basis for thetopology of X for x ∈ p−1(y) ⊂ p−1(V ), there is some U ∈ B which is an open neighborhoodof x contained in p−1(V ). It follows that p(U) ⊂ V is an open neighborhood of y, provingthat {p(U) | U ∈ B} is in fact a basis for the topology of X/G.

To show that X/G is Hausdorff, let y, y′ ∈ X/G with y 6= y′, and pick points x ∈ p−1(y),x′ ∈ p−1(y′). Proceeding like in part (a) the points gx, g′x′ for g, g′ ∈ G are distinct pointsof X. Hence there are mutually disjoint open neighborhoods Ugx, Ug′x′ of these points. Itfollows that

U :=⋂g∈G

g−1Ugx U ′ :=⋂g′∈G

(g′)−1Ug′x′

are open neighborhoods of x resp. x′. Moreover their image under p are disjoint openneighborhoods of y resp. y′ since p−1(p(U) and p−1(p(U ′)) are disjoint. This proves thatX/G is Hausdorff.

Part (c). To show that the Z/k action on S2n−1 is free, assume [m] ∈ Z/k belongs to theisotropy subgroup of some v = (v1, . . . , vn) ∈ S2n−1, that is, e2πim/kv = v. Since v is a unitvector, it has a non-zero component vi ∈ C. Then e2πim/kvi = vi which implies e2πim/k = 1,which in turn implies that m ∈ Z must be a multiple of k. In particular, [m] is the identityelement in Z/k, proving that the action is free.

5. Let p : X → X be a covering map. Assume that X is path connected, let x0 ∈ X and letx0 = p(x0) ∈ X.

(a) Show that the kernel of the induced homomorphism p∗ : π1(X, x0)→ π1(X, x0) is trivial.Hint: use the lifting property for families of paths.

29

(b) Show that the map

Φ: π1(X, x0) −→ p−1(x0) defined by [γ] 7→ γ(1)

is a well-defined surjection. Here γ : I → X is the lift of γ with γ(0) = x0.

(c) Show that Φ(g1) = Φ(g2) for g1, g2 ∈ G := π1(X, x0) if and only if Hg1 = Hg2, where

Hgi ⊂ G are right cosets for the subgroup H = p∗(π1(X, x0)) ⊆ G.

We note that putting parts (b) and (c) together, we obtain a bijection between the set ofright cosets H\G and the fiber p−1(x0).

Proof. Part (a). Let γ : I → X be a based loop in X representing an element in the kernelof p∗. This means that the loop p ◦ γ is homotopic to the constant loop cx0 via a homotopyH : I × I → X. Writing this out more explicitly we have

H(s, 0) = p(γ(s)) H(s, 1) = x0 = H(0, t) = H(1, t)

In particular, the first condition implies that the diagram

I × {0} γ //

��

X

p

��I × I H // X

commutes, and hence we can use the lifting property for the covering space X → X to obtaina map H as shown in the diagram

I × {0} γ //

��

X

p

��I × I H //

H

;;xxxxxxxxxxX

We note that for H(0, t) ∈ p−1(x0), since H is a lift of H and H has the property H(0, t) = x0.Since the map

I → p−1(x0) ⊂ X given by t 7→ H(0, t)

is a continuous map from a connected space to a discrete space, it must be constant. SinceH(0, 0) = γ(0) = x0 we must in fact have H(0, t) = x0 for all t ∈ I. Similarly, we can show

that H(1, t) = x0 for all t ∈ I. Applying the same argument to the map

I → p−1(x0) ⊂ X given by s 7→ H(s, 1)

we conclude that H(s, 1) = x0. These properties of H show that H is a homotopy betweenthe loop γ and the constant loop cx0 , which implies that [γ] is the identity element in

π1(X, x0).

30

Part (b). To show that Φ is well-defined, let γ, γ′ : I → X be loops in X based at x0

representing the same element in π1(X, x0). Then we can interpret a homotopy H : I×I → Xbetween these loops as a family of loops parametrized by the point s ∈ I, that is

H(0, t) = γ(t) H(1, t) = γ′(t) H(s, 0) = x0 = H(s, 1).

Let H be the unique lift of this homotopy determined by the commutative diagram

I × {0}cx0 //

��

X

p

��I × I

H

;;wwwwwwwwwwH // X,

where cx0 is the constant map to the point x0 ∈ X. Restricting to s = 0, 1 we obtain paths

γ(t) := H(0, t) and γ′(t) := H(1, t) which are the unique lifts of the loops γ, γ′ to paths in

X starting at x0.Now consider the path I → X given by s 7→ H(s, 1). Since H(s, 1) = x0 for all s, this is

a path in the discrete space p−1(x0), and hence as in part (a) we conclude that

γ(1) = H(0, 1) = H(1, 1) = γ′(1)

which shows that the map Φ is well-defined.Surjectivity of Φ is easy: given any point x ∈ p−1(x0) let γ : I → X be a path with

γ(0) = x0 and γ(1) = x. Then γ := p ◦ γ is a loop in X based at x0, and γ is the unique liftof γ with starting point x0. This shows that Φ maps [γ] ∈ π1(X, x0) to γ(1) = x.

Part (c). Suppose that Φ(g1) = Φ(g2) for elements gi ∈ G = π1(X, x0). Representing gi by

loops γi in X based at x0, this means that γ1(1) = γ2(1), where γi : I → X is the unique liftof γi with γi = x0. In particular, the concatenation γ1 ∗ γ2 is defined, and is a loop in Xbased at x0. Then

p∗([γ1 ∗ γ2]) = [(p ◦ γ1) ∗ (p ◦ γ2)] = [γ1 ∗ γ2] = g1g−12

is an element of H = p∗(π1(X, x0)) which implies Hg1 = Hg2.Conversely, assume Hg1 = Hg2. Then g2 = hg1 for some element h ∈ H. Let g1 ∈

π1(X, x0) be represented by the based loop γ1, and let γ1 be the unique lift of this loop to a

path in X starting at x0. Let h ∈ π1(X, x0) be the unique element mapping to h ∈ π1(X, x0)

via p∗, and let δ be a based loop in X representing h. Then δ := p ◦ δ is a based loop inX representing h ∈ π1(X, x0). We note that γ2 := δ ∗ γ1 is a loop representing g2, and that

γ2 = δ ∗ γ1 is the unique lift of this loop (since δ is a loop). By part (b) this implies

Φ(g2) = γ2(1) = (δ ∗ γ1)(1) = γ1(1) = Φ(g1).

31


1. Let p : (E, e0)→ (B, b0) be a covering space, and let f : (X, x0)→ (B, b0) be a map withX path-connected and locally path-connected. Show:

(a) A lift f : (X, x0)→ (E, e0) of f exists if and only if f∗(π1(X, x0)) ⊆ p∗(π1(E, e0)). Hint:use the path lifting property for the covering space E → B.

(b) There is at most one such lift.

Proof. Part (a). Suppose that a lift f : (X, x0) → (E, e0) of f exists. This means that wehave the following commutative diagram of pointed spaces and basepoint preserving maps:

(E, e0)

p

��(X, x0)

f99ttttttttt

f// (B, b0)

Applying the fundamental group functor results in the following commutative diagram inthe category of groups:

π1(E, e0)

p∗��

π1(X, x0)

f∗88qqqqqqqqqqq

f∗// π1(B, b0)

This shows that f∗(π1(X, x0)) is contained in p∗(π1(E, e0)).Conversely, let us assume f∗(π1(X, x0)) ⊆ p∗(π1(E, e0)). To construct the desired lift

f : (X, x0) → (E, e0) we choose for a given point x ∈ X a path γ : I → X with γ(0) = x0

and γ(1) = x; such a path exists due to the assumption that X is path-connected. Applyingthe path lifting property for the covering space E → B to the path δ := f ◦ γ : I → B, thereis a unique lift δ : I → E of δ with δ(0) = e0. More explicitly, we apply the homotopy liftingproperty to the diagram

{0}

��

e0 // E

p

��I

δ

66

δ

77nnnnnnnn γ // Xf // B

Then we define f(x) := δ(1) ∈ E. It remains to show:

(i) the map f : X → E is well-defined and basepoint preserving.

(ii) f is continuous.

For (i), let us assume that γ′ is another path from the basepoint x0 to the given point x ∈ X,

let δ′ := f ◦ γ′, and let δ′ : I → E be a lift of δ′ with δ′(0) = e0. Then the concatenationγ ∗ γ′ is a based loop in (X, x0) and its image

f ◦ (γ ∗ γ′) = (f ◦ γ) ∗ (f ◦ γ′) = δ ∗ δ′

32

is a based loop in (B, b0). By assumption, the element f∗([γ ∗ γ′]) = [δ ∗ δ′] ∈ π1(B, b0) is inthe image of p∗ : π1(E, e0)→ π1(B, b0).

According to a proposition in class a based loop α : I → B represents an element in theimage of p∗ : π1(E, e0)→ π1(B, b0) if and only if the unique lift α : I → E with α(0) = e0 is

a loop. Applying this to the loop α = δ ∗ δ′ shows that the lifts δ and δ′ of δ, δ′ have thesame endpoint: restricting the loop α to the two subintervals [0, 1/2] and [1/2, 1] we write αas the concatenation of two paths, which will be lifts of the two paths δ, δ′ that α = δ ∗ δ′ is

built from. Then uniqueness of pathlifting implies that these paths must be δ and δ′. Thefact that α is a loop then implies that the two endpoints of the paths δ and δ′ must agree.

The map f is basepoint preserving by construction (for x = x0 we choose γ to be the

constant path at x0, and then δ, δ will be the constant paths at b0 and e0, respectively.To prove continuity of f , we note that continuity of a map is a local property, i.e., it

suffices to prove continuity of f restricted to a sufficiently small open neighborhood V ofevery point x ∈ X. Given x ∈ X we construct V as follows: the point f(x) ∈ B has aneighborhood U such that p−1(U) is a union of disjoint open subsets Ui ⊂ E, i ∈ I, suchthat the restriction p|Ui : Ui → U is a homeomorphism for each i ∈ I (by the definition ofa covering map). By continuity of f the preimage f−1(U) is then an open neighborhood ofx ∈ X. The assumption that X is locally path-connected implies that there is a possiblysmaller open neighborhood V ⊂ f−1 of x which is path connected.

We claim that if f(x) belongs to Ui, then f(y) ∈ Ui for every y ∈ V . To see this, suppose

as above that γ is a path in X from x0 to x, and δ is a path in E starting at e0 which is alift of f ◦ γ. Then f(x) = δ(1). To determine f(x′) for a point x′ ∈ V we choose γ′ = ε ∗ γ,

where ε is a path in V starting at x and ending in x′. Then by construction f(x′) is the

endpoint of the lift of δ′ := (f ◦ ε) ∗ (f ◦ γ) = (f ◦ ε) ∗ δ. This is the concatenation of δ

(which goes from e0 to f(x)) and of a lift β of the path (f ◦ ε) with β(0) = f(x). The

homeomorphism p|Ui : Ui≈−→ U restricts to a homeomorphisms between Vi := Ui ∩ p−1(V )

and V , which implies that Vi is path-connected, since V is. In particular, since β(0) = f(x)

belongs to Vi, so does the endpoint β(1) = f(x′). This proves the claim above.We note that the composition

Xf // E

p // B

is the map f which is contininuous by assumption. This does not imply that the factorf is continuous (think of a counterexample in the case where X is the unit interval I and

f : I → E is a path in E). However, upon restriction of f to V the factorization takes thefollowing form:

V

f|V

77f|V // Ui

p // U .

Since p : Ui → U is a homeomorphism, the continuity of f|V does then indeed imply the

continuity of f|V .

Part (b). If f , : (X, x0)→ (E, e0) is a lift of f , the value f(x) ∈ E can be described as in

part (a) as f(x) = δ(1), where we chose a path γ from x0 to x and a lift δ of the image path

f ◦ δ. The uniqueness of path-lifting then implies the uniqueness of the lift f .

33

2. Show that if a path-connected, locally path-connected space X has finite fundamentalgroup, then every map X → S1 is homotopic to a constant map. Hint: Use the coveringspace R→ S1 and the previous homework problem.

Proof. Let f∗ : π1(X, x0) → π1(S1, y0), y0 := f(x0) be the homomorphism of fundamentalgroups induced by the map f : X → S1. We note that every element a ∈ π1(X, x0) hasfinite order and hence f∗(a) ∈ π1(S1) ∼= Z has finite order which implies that f∗(a) = 0.In particular f∗(π1(X, x0)) is contained in the image of p∗ : π1(R) → π1(S1) induced by thecovering map p : R→ S1. By a result proved in class, this implies that f factors in the formf = p ◦ f for some map f : X → R. The map f is homotopic to the constant map sendingevery point x ∈ X to y0 := f(x0) via the linear homotopy

H : X × I −→ R given by H(x, t) = tf(x) + (1− t)y0

Then H := p ◦ H : X × I → S1 is a homotopy between f and the constant map at y0.

3. Let U± := Sn \ {(±1, 0, . . . , 0)}. We recall that the stereographic projection is the mapP± : U± → Rn given by P±(x0, x1, . . . , xn) = 1

1∓x0 (x1, . . . , xn).

(a) Show that (U+, P+) and (U−, P−) are charts for Sn.

(b) Show that {(U+, P+), (U−, P−)} is a smooth atlas for Sn.

(c) Show that the atlas above is smoothly compatible with the smooth atlas we’ve dis-cussed in class, consisting of the charts (U±i , φ

±i ), where U±i ⊂ Sn consists of the vectors

(x0, . . . , xn) ∈ Sn such that xi > 0 resp. xi < 0, and φ±i (x0, . . . , xn) = (x0, . . . , xi, . . . , xn).

Proof. Part (a). Let ε ∈ {±1}. We need to show that the map Pε : Uε → Rn is a homeo-morphism. To do this, we calculate the inverse of Pε explicitly (we need the explicit form ofthe inverse for later parts anyway). So, given y = (y1, . . . , yn) ∈ Rn we want to show thatthere is a unique x = (x0, x1, . . . , xn) ∈ Uε = Sn \ (−ε, 0, . . . , 0) satisfying the equation

(y1, . . . , yn) = Pε(x0, . . . , xn) =1

1− εx0

(x1, . . . , xn). (7.1)

Taking the norm squared of both sides we obtain

||y||2 =x2

1 + · · ·+ x2n

(1− εx0)2=||x||2 − x2

0

(1− εx0)2=

1− x20

(1− εx0)2.

This leads to the quadratic equation

||y||2(1− εx0)2 = 1− x20,

for x0. In standard form this equation becomes

(||y||2 + 1)x20 − 2ε||y||2x0 + (||y||2 − 1) = 0

34

and hence we have solutions

x0 =2ε||y||2 ±

√4||y||4 − 4(||y||2 + 1)(||y||2 − 1)

2(||y||2 + 1)=

2ε||y||2 ± 2

2(||y||2 + 1)=ε||y||2 ± 1

||y||2 + 1.

We observe that we don’t want the solution x0 = ε||y||2+ε||y||2+1

= ε, since that implies xi = 0 for

1 ≤ i ≤ n, and hence forces x = (ε, 0, . . . , 0) which does not belong to Uε. This forces

x0 =ε(||y||2 − 1)

||y||2 + 1.

Using equation (7.1) this implies

xi = yi(1− εx0) = yi(1− εε(||y||2 − 1)

||y||2 + 1) = yi(1−

||y||2 − 1

||y||2 + 1) = yi

2

||y||2 + 1

These considerations imply that the inverse map P−1ε : Rn −→ Uε is given by

P−1ε (y1, . . . , yn) =

1

||y||2 + 1

(ε(||y||2 − 1), 2y1, . . . , 2yn

).

Since all components of P−1ε are continuous functions, this is a continuous map and hence

Pε is a homeomorphism.

Part (b). We need to show that the transition map

Pε(Uε ∩ U−ε)P−1ε // Uε ∩ U−ε

P−ε // P−ε(Uε ∩ U−ε)

is a smooth map for ε = ±1. We note that Uε ∩ U−ε = Sn \ {±1, 0, . . . , 0} and hence

P±ε(Uε ∩ U−ε) = Rn \ {0}.

Explicitly, for y = (y1, . . . , yn) ∈ Rn \ {0} we have

P−ε(P−1ε (y)) = P−ε(

1

||y||2 + 1

(ε(||y||2 − 1), 2y1, . . . , 2yn

))

=1

1 + ε ε(||y||2−1)

||y||2+1

(2y1

||y||2 + 1, . . . ,

2yn||y||2 + 1

)=

1

||y||2 + 1 + (||y||2 − 1)(2y1, . . . , 2yn)

=1

||y||2(y1, . . . , yn)

This map is smooth, since all its components are smooth functions (the function 1/||y||2 issmooth on Rn \ {0}, the domain of P−ε ◦ P−1

ε ).

Part (c). For δ = ±1 the map φδi : U δi → Dn is given by forgetting the i-th coordinate.

Hence the composition φδi ◦P−1ε is a smooth function, since all components of P−1

ε are smooth.

35

To check smoothness of the composition Pε ◦ (φδi )−1 we recall that (φδi )

−1 : Dn −→ U δi is

given by(φδi )

−1(y1, . . . , yn) = (y1, . . . , yi−1, δ√

1− ||y||2, yi, . . . , yn).

It follows that

Pε ◦ (φδi )−1(y) =

1

1−εδ√

1−||y||2(y1, . . . , yn) i = 0

11−εy1 (y2, . . . , yi−1, δ

√1− ||y||2, yi, . . . , yn) i 6= 0

We note that there are two potential problems with the smoothness of these maps:

• The square root function is not differentiable at 0; so there is a potential issue when1 − ||y||2, the argument of the square root, becomes zero. This doesn’t happen sincey ∈ D2, the interior of the 2-disk.

• The function is undefined if the denominator becomes 0. We could discuss for which ythis happens in either case (i = 0 or i 6= 0) and show that those y are not in the domainof the transition map. It is easier to say that by construction the transition map iswell-defined on its domain, and hence we don’t need to worry about those points.

4. Let Ui := {[z0, z1, . . . , zn] ∈ CPn | zi 6= 0} for i = 0, . . . , n and let

φi : Ui −→ Cn be defined by φi([z0, z1, . . . , zn]) =

(z0

zi, . . . ,

zizi, . . . ,

znzi

).

(a) Show that (Ui, φi) is a chart for CPn.

(b) Show that {(Ui, φi) | i = 0, . . . , n} is a smooth atlas for CPn.

Proof. Part (a). It is easy to check that the inverse of φi : Ui −→ Cn is given by

φ−1i (w1, . . . , wn) = [w1, . . . , wi, 1, wi+1, . . . , wn] ∈ Ui for (w1, . . . , wn) ∈ Cn.

This map is the composition of the map

Cn → Cn+1 \ {0} given by (w1, . . . , wn) 7→ (w1, . . . , wi, 1, wi+1, . . . , wn)

composed with the projection map p : Cn+1\{0} → ((Cn+1 \ {0})/ ∼) = CPn. The first mapis continuous since its components are, projection maps onto quotient spaces are continuousby the construction of the quotient topology and hence the composition φ−1

i is continuous.This shows that φi : Ui → Cn is a homeomorphism. In other words, (Ui, φi) is a chart forCPn.

Part (b). The Ui’s form an open cover of CPn and so the collection (Ui, φi) is an atlas forCPn. It remains to show that it is a smooth atlas, i.e., that the transition map

Cn ⊃ φi(Ui ∩ Uj)φ−1i // Ui ∩ Uj

φj // φj(Ui ∩ Uj) ⊂ Cn

36

is smooth for all i, j = 0, . . . , n. For w = (w1, . . . , wn) ∈ φi(Ui ∩ Uj), i 6= j we compute

φj(φ−1i (w) = φj([w1, . . . , wi, 1, wi+1, . . . , wn]

=

(

w1

wj+1, . . . ,

wj+1

wj+1, . . . , wi

wj+1, 1wj+1

, . . . , wnwj+1

)0 ≤ j < i(

w1

wj, . . . , wi

wj, 1wj, wi+1

wj, . . . ,

wjwj, . . . , wn

wj

)i < j ≤ n

As in the previous homework problem this shows that the transition map is smooth since itscomponents are all smooth functions (since the transition function is well-defined those w’sfor which the denominator is 0 can’t be in the domain of the transition function).

5. Show that the Cartesian product of M ×N of smooth manifolds of dimension m resp. nis again a smooth manifold of dimension m+ n.

Proof. Let (Ui, φi)i∈I be a smooth atlas for M , and (Vj, ψj)j∈J for N . We claim that thecollection of charts of M ×N given by

M ×N ⊃open

Ui × Vjφi×ψj // φi(Ui)× ψj(Vj) ⊂

openRm × Rn = Rm+n

for (i, j) ∈ I × J is a smooth atlas for M ×N . In particular, M ×N is a smooth manifoldof dimension m+ n.

It is clear that φi × ψj is a homeomorphism with inverse given by φ−1i × ψ−1

j . Also it isclear that the open subsets Ui × Vj cover M × N . Hence this collection is an atlas, and itonly remains to show that this is a smooth atlas.

The transition map between the charts φi×ψj and φi′ ×ψj′ is given by (after restrictingthe domains in the obvious way) by

(φi′ × ψj′) ◦ (φi × ψj)−1 = (φi′ × ψj′) ◦ (φ−1i × ψ−1

j ) = (φi′ ◦ φ−1i )× (ψj′ ◦ ψ−1

j )

The map φi′ ◦ φ−1i is a transition map for a smooth atlas for M and hence a smooth map

between open subsets of Rm. Similarly the map ψj′ ◦ ψ−1j is a smooth map between open

subsets of Rn. It follows that the Cartesian products of these maps is a smooth map (betweenopen subsets of Rm×Rn = Rm+n) and hence the atlas constructed above is in fact smooth.


1.

(a) Let M be a smooth manifold and let f : N →M be a homeomorphism. Show that thereis a unique smooth structure on N such that f is a diffeomorphism.

(b) Give the space {x ∈ Rn+1 | max{|xi|} = 1} (the boundary of the (n + 1)-cube) thestructure of a smooth manifold.

(c) Applying part (a) to N = {(x, y) ∈ R2 | y2 = x3} and the homeomorphism f : N →M = R given by f(x, y) = y gives N the structure of a smooth manifold. Is the inclusionmap i : N → R2 a smooth map?

37

Proof. Part (a). Let (hα : Uα → U ′α)α∈A be the maximal atlas giving the smooth structureon M . For α ∈ A, let Vα := f−1(Uα) (which is an open subset of N), and let gα be thecomposition

Vαf

≈//Uα

hα

≈//U ′α

We claim that

1. (gα : Vα → U ′α)α∈A is a smooth atlas for N .

2. This atlas is maximal.

3. With this smooth structure on N , the homeomorphism f is a diffeomorphism.

4. If (g : V → U ′) is a chart belonging to a maximal atlas of N such that f is a diffeo-morphism, then g = gα for some α ∈ A.

Claims 1, 2 and 3 show that there exists a smooth structure on N such that f is smooth.Claim 4 shows that any smooth structure on N such that f is a diffeomorphism is equal tothe smooth structure constructed above; in particular, there is a unique smooth structureon N such that f is a diffeomorphism, proving part (a).

To prove that (gα : Vα → U ′α)α∈A is a smooth atlas we consider the transition map gβ ◦g−1α

relating the charts gα and gβ for α, β ∈ A. The following commutative diagram shows thatthis transition map is equal to the transition map hβ ◦ h−1

α and hence is smooth:

Vα ∩ Vβgα

≈sshhhhhhhhh

hhhhhhhh

hhh

f ≈

��

gβ

≈ ++VVVVVVVV

VVVVVVVV

VVVV

gα(Vα ∩ Vβ) = hα(Uα ∩ Uβ) hβ(Uα ∩ Uβ) = gβ(Vα ∩ Vβ)

Uα ∩ Uβhα

≈kkVVVVVVVVVVVVVVVVVVVV hβ

≈44hhhhhhhhhhhhhhhhhhhh

To prove that the smooth atlas (gα)α∈A is maximal we need to show that if g0 : V0 → V ′0is a chart for N such that all transition maps gα ◦ g−1

0 for charts (gα)α∈A are smooth, theng0 is equal to the chart gα for some α ∈ A. To show this, let h0 be the composition

U0 := f(V0)f−1

≈//V0

g0

≈//U ′0

Then the diagram above for β = 0 shows that the transition maps h0 ◦ h−1α and hα ◦ h−1

0 forany α ∈ A are smooth. It follows that the atlas (hα)α∈Aq{0} is a smooth atlas; by maximality,the chart h0 must be equal to hα0 for some α0 ∈ A. It follows that the chart g0 is equal togα0 and hence the atlas (gα)α∈A is a maximal smooth atlas.

To prove claim 4 we note that the assumptions imply that hα ◦ f ◦ g−1 = gα ◦ g−1 andg ◦ f−1 ◦ hα = g ◦ g−1

α are smooth. In other words, the chart (V, g) is smoothly compatiblewith any of the charts (Vα, gα), α ∈ A. The maximality of the atlas (gα)α∈A then impliesg = gβ for some β ∈ A.

38

Part (b). To prove part (b) we note that the map

X = {x ∈ Rn+1 | max{|xi|} = 1} f−→ Sn defined by f(x) =x

||x||

is continuous. It is a homeomorphism since its inverse can be described explicitly as f−1(x) =x

||x||maxwhere ||x||max = max{|xi|}, which shows in particular that f−1 is continuous (the map

Rn → R, x 7→ ||x||max is continuous thanks to the estimate ||x||max ≤ ||x||). Then the resultsof part (a) allow us to use this homeomorphism to equip X with the structure of a smoothmanifold.

Part (c). In general for a smooth n-manifold N a map h : N → Rk is smooth if and only ifthe composition

U ′ϕ−1// U ⊂ N

h // Rk

is smooth for every chart N ⊃ Uϕ→ U ′ ⊂ Rn in the smooth atlas defining the smooth

structure on N . For the manifold N at hand, the smooth structure on N constructed by the

method of part (a) is given by the atlas consisting of the single chart N = Uf→ U ′ = R. So

i is smooth if and only if the composition

i ◦ f−1 : R // R2 given by y 7→ (y2/3, y)

is smooth. This shows that the inclusion map i is not smooth, since the first component ofthe map i ◦ f−1 is not smooth (the derivative of y2/3 is −y−1/3 for y 6= 0, but does not existfor y = 0).

2. Show that Sn ⊂ Rn+1 is a smooth manifold by proving that 1 is a regular value of thesmooth function f : Rn+1 → R, x 7→ ||x||2.

Proof. We need to show that for every x ∈ Sn = f−1(1) the differential

(Df)x : Rn+1 −→ R

is surjective. As in class we calculate (Df)x via (Df)x(y) = ∂f(x+sy)∂s

(0).

f(x+ sy) = ||x+ sy||2 = 〈x+ sy, x+ sy〉 = 〈x, x〉+ 2s〈x, y〉+ s2〈y, y〉

and hence

(Df)x(y) =∂f(x+ sy)

∂s(0) = 2〈x, y〉

Taking y = x we see that (Df)x is a surjective linear map for x ∈ f−1(1).

3. Let Mn×k(R) be the vector space of n× k-matrices. For A ∈Mn×k(R) let At ∈Mk×n(R)be the transpose of A, and let Sym(Rk) = {B ∈ Mk×k(R) | Bt = B} be the vector space ofsymmetric k × k-matrices.

(a) Show that the map Φ: Mn×k(R)→ Sym(Rk), A 7→ AtA is smooth.

(b) Show that the identity matrix is a regular value of the map Φ.

39

(c) Let Vk(Rn) ⊂ Mn×k(R) be the subset consisting of those n × k matrices whose columnvectors are unit length and pairwise perpendicular. Show that Vk(Rn) is a smoothmanifold of dimension kn− k(k + 1)/2. Hint: what is the relationship between Vk(Rn)and the map Φ?

We remark that identifying Mn×k(R) in the usual way with the vector space Hom(Rk,Rn)of linear maps f : Rk → Rn, a matrix belongs to Vk(Rn) if and only if the correspondinglinear map f is an isometry, that is, if f preserves the length of vectors in the sense that||f(v)|| = ||v||, or equivalently, if f preserves the scalar product in the sense that

〈f(v), f(w)〉 = 〈v, w〉 for all v, w ∈ Rk.

The manifold Vk(Rn) is called the Stiefel manifold. We observe that Vn(Rn) is the orthogonalgroup O(n) of isometries Rn → Rn. In particular, this problem shows that O(n) is a manifoldof dimension n2 − n(n+ 1)/2 = n(n− 1)/2.

Proof. The entries of the matrix AtA are quadratic polynomials in the entries of A and hencethe map Φ is smooth.

To prove part (b) we first calculate the differential (DΦ)A(C), C ∈ Mn×k(R) as in theprevious problem as the derivative of the function Φ(A+ sC) at s = 0.

Φ(A+ sC) = (A+ sC)t(A+ sC) = AtA+ s(CtA+ AtC) + s2CtC

and hence∂Φ(A+ sC)

∂s(0) = CtA+ AtC

To show that for A ∈ Φ−1(I) (I ∈Mk×k(R) is the identity matrix) the differential

(DΦ)A : Mn×k(R)→ Sym(Rk)

is surjective, let B ∈ Sym(Rk) and let C = AB ∈Mn×k(R). Then

(DΦ)A(C) = CtA+ AtC = (AB)tA+ At(AB) = BtAtA+ AtAB = Bt +B = 2B,

which proves surjectivity.According to a theorem proved in class, if f : V → W is a smooth map between finite

dimensional real vector spaces, then the preimage f−1(w) of a regular value w ∈ W is amanifold of dimension dimV −dimW . Parts (a) and (b) show that this theorem is applicablein our situation and hence Vk(Rn) is a manifold of dimension

dimMn×k(R)− dim Sym(Rk)

It remains to show that

dimMn×k(R) = kn and dim Sym(Rk) = k(k + 1)/2

The first statement follows from the fact that a basis of Mn×k(R) is provided by the matrices{A(i, j)}, 1 ≤ i ≤ n, 1 ≤ j ≤ k, whose entries are all zero except the ij-th entry which is1. There are nk matrices A(i, j) and hence dimMn×n(R) = nk. Similarly, a basis for the

40

symmetric k × k matrices is given by the matrices {B(i, j)}, 1 ≤ i ≤ j ≤ k; here B(i, j) isthe matrix whose entries are zero except the ij-th entry and the ji-th entry which are both 1(obviously, these are the same entries for i = j). For a fixed j between 1 and n, the numberof matrices B(i, j) with 1 ≤ k ≤ j is j. It follows that

dim Sym(Rk) = 1 + 2 + 3 + · · ·+ k = k(k + 1)/2

4. Let M , N be non-empty smooth manifolds of dimension m resp. n. Show that if M isdiffeomorphic to N , then m = n.

Proof. Let f : M → N be a diffeomorphism and x ∈M . Let φ : U → U ′ ⊂ Rm be a smoothchart for M with x ∈ U , and let ψ : V → V ′ ⊂ Rn be a smooth chart for N with f(x) ∈ V .Replacing U , V by smaller open subsets, we can assume that f restricts to a diffeomorphismfrom U to V . Then the composition

Rm ⊃open

U ′φ−1// U

f|U // Vψ // V ′ ⊂

openRn

is a diffeomorphism g between U ′ ⊂ Rm and V ′ ⊂ Rn. The differential of g at any pointy ∈ U ′ is then an isomorphism from Rm to Rn (whose inverse is the differential of g−1 at thepoint g(y)). This implies m = n.

9 Homework Assignment #9

1. Let M1, M2 be smooth manifolds, and let πj : M1 ×M2 →Mj be the projection onto thej-th factor for j = 1, 2. Show that for any (p1, p2) ∈M1 ×M2 the map

α : T(p1,p2)(M1 ×M2) −→ Tp1M1 ⊕ Tp2M2

defined byα(v) = (Tp1π1(v), Tp2π2(v))

is an isomorphism. Remark: Using this isomorphism, we will routinely identify TpjMj witha subspace of T(p1,p2)(M1 ×M2).

Proof. For ` = 1, 2 let i` : M` → M1 ×M2 be the inclusion map given by i1(x1) = (x1, p2),i2(x2y) = (p1, x2). Then for `, j = 1, 2 the composition

M`i`−→M1 ×M2

πj−→Mj

is the identity map for ` = j, and the constant map x` 7→ pj otherwise. Differentiating it, itfollows that the composition

Tp`M`(i`)∗−→ T(p1,p2)(M1 ×M2)

(πj)∗−→ TpjMj

41

is the identity map for ` = j and the trivial map otherwise. It follows that the composition

Tp`M`(i`)∗−→ T(p1,p2)(M1 ×M2)

α−→ Tp1M1 ⊕ Tp2M2

sends a vector X` ∈ Tp`M` to (X, 0) for ` = 1 and (0, X) for ` = 2. Since every pair (X1, X2)can be written as a sum of pairs of this type, the map α is surjective. This implies that α isan isomorphism, since the dimension of domain and range is both dimM1 + dimM2.

2. Let M , W be smooth manifolds of dimension dimM = n, dimW = k, and let f : M → Wbe a smooth map. We recall that w ∈ W is a regular value of f if every x ∈ f−1(w) is aregular point of f , i.e., the differential Txf : TxM → TwW is surjective.

(a) Show that N = f−1(w) is a smooth submanifold of W of dimension n− k. Hint: reduceto the situation proved in class where M , W are open subsets of Euclidean space.

(b) Show that for x ∈ N = f−1(w) the tangent space TxN ⊂ TxM is the kernel of thedifferential f∗ : TxM → TwW .

Proof. Part (a). Given a point x ∈M := f−1(p), we need to find a chart N ⊃ Uϕ−→ U ′ ⊂

Rn such that ϕ(U ∩M) = U ′ ∩ Rn−k. Let P ⊃ Wχ→ W ′ ⊂ Rk be a chart with p ∈ W , and

let N ⊃ Vψ→ V ′ ⊂ Rn be a chart with x ∈ V . Choosing a smaller open neighborhood V of

x if necessary, we can assume that the smooth map f restricts to a smooth map f : V → W .Identifying V (resp. W ) with the open subsets V ′ (resp. W ′) of Euclidean space via thediffeomorphisms ψ (resp. χ), we can apply our theorem from class according to which thelevel set f−1(p) ⊂ V of the regular value p ∈ W is a submanifold of dimension n− k, givingus in particular the desired submanifold chart (U,ϕ) at the point x ∈ V .

Part (b). To show that i∗ is injective consider the following commutative diagram of smoothmaps

U ∩Mi��

ϕ|U∩M // U ′ ∩ Rn−k ⊂ Rn−k

j��

Uϕ // U ′ ⊂ Rn

Here i′ : U ∩M → U ⊂ N is the restriction of the inclusion map i : M ↪→ N to the opensubset U ∩M ; since i and i′ agree in a neighborhood of x ∈M , we have the equality i∗ = i′∗for their differentials. The map j is the inclusion of Rn−k in Rn given by j(v1, . . . , vn−k) =(v1, . . . , vn−k, 0, . . . , 0). The horizontal maps are charts for M resp. N containing the point x.In particular, these a local diffeomorphisms, whose differentials are therefore isomorphismsof tangent spaces. Functoriality of the derivative then leads to the following commutativediagram of linear maps

TxM(ϕU∩M )∗∼=

//

i∗��

Tϕ(x)Rn−k = Rn−k

j∗

��TxN

ϕ∗∼=

// Tϕ(x)Rn = Rn

42

This shows that the map i∗ is injective, since it corresponds via the horizontal isomorphismsof the above diagram to the injective map j∗ = j (the derivative of the linear map j is againj).

To show that the image of i∗ : TxM → TxN is equal to the kernel of f∗ : TxN → TpP , wenote that the composition

M i // Nf // P

is the constant map that sends every point of M = f−1(p) to p. This implies that thederivative of f ◦ i is zero. Since (f ◦ i)∗ = f∗ ◦ i∗, this implies that the image of i∗ iscontained in the kernel of f∗. To prove im i∗ = ker f∗, we will argue that these vectorspaces have the same dimension, and hence im i∗ ⊂ ker f∗ implies their equality. Since i∗ isinjective, dim im i∗ = dimTxM = dimM = n− k. Since f∗ : TxN → TpP is surjective by theassumption that p is a regular value of f , we have

dim ker f∗ = dimTxN − dim im f∗ = dimTxN − dimTpP = n− k.

3. Let O(n) be the orthogonal group (which is the submanifold of Mn×n(R) consisting ofthose n × n matrices A with AtA = e; here e denotes the n × n identity matrix which isthe identity element of the group O(n)). Identify the tangent space TeO(n) as a subspace ofTeMn×n(R) = Mn×n(R). Hint: Use part (b) of the previous problem.

Proof. Let Φ: Mn×n(R)→Mn×n(R) be the map given by A 7→ AtA. Then by definition theorthogonal group O(n) is the level set Φ−1(11) where 11 ∈Mn×n(R) is the identity matrix. Weremark that 11 can’t possibly be a regular value of the smooth map f : Mn×n(R)→Mn×n(R),since otherwise O(n) = f−1(11) would be a submanifold of Mn×n(R) whose dimension is thedifference of the dimensions of the vector spaces that are the domain resp. range of f , leadingto the obviously wrong conclusion that O(n) is a manifold of dimension 0!

To fix this problem, we note that AtA is always a symmetric matrix, since (AtA)t =At(At)t = AtA. This allows us to think of Φ as a smooth map

Φ: Mn×n(R) −→ Sym(Rn)

to the vectorspace Sym(Rn) of symmetric n× n-matrices.In problem 3 of the last homework set we computed the differential of Φ (actually, the

map Φ of that problem is more general, but specializing to k = n it agrees with the mapconsidered here). We showed that the differential

(DΦ)A : TAMn×n(R) −→ TΦ(A)Sym(Rn) is given by (DΦ)A(C) = CtA+ AtC,

and went on to argue that (DΦ)A is surjective for A ∈ Φ−1(11) = O(n) which shows thatO(n) is a submanifold of Mn×n(R).

By part (b) of problem 2 of this set, the tangent space of a level set Φ−1(λ) of a regularvalue λ at a point A ∈ Φ−1(λ) is equal to the kernel of the differential (DΦ)A. In the caseat hand, this means that the tangent space TIO(n) at the identity element e = I ∈ O(n) isequal to the kernel of the differential

(Dφ)I : TIMn×n(R) = Mn×n(R) −→ TISym(Rn) = Sym(Rn)

43

which we computed to be (DΦ)I(C) = Ct+C. This shows that TIO(n) = ker(DΦ)I consistsof the vector space of skew-symmetric matrices, that is, matrices C satisfying Ct = −C.

4. Show that the projection map p : S2n+1 → CPn is a submersion, i.e., the differentialp∗ : TzS

2n+1 → Tp(z)CPn is surjective for every point z ∈ S2n+1. (In particular, by 2(a) eachfiber p−1(L), L ∈ CPn is a submanifold of dimension 1). Hint: The projection map p extendsto a projection map P : Cn+1 \ {0} → CPn. Argue first that P is a submersion, using theobvious identification of tangent spaces of the domain with Cn+1, and using the differentialof our standard charts for CPn to identify the tangent spaces of the range with Cn.

Proof. Let z = (z0, . . . , zn) ∈ S2n+1 ⊂ Cn+1, and let p(z) = [z0, . . . , zn] ∈ CPn be the imageof z under the projection map. We recall from class that the tangent space TzS

2n+1 can beidentified with the vectors w = (w0, . . . , wn) ∈ Cn which are perpendicular to z, i.e., thosew for which the Hermitian inner product

〈z, w〉 = z0w0 + z1w1 + · · ·+ znwn

vanishes. This suggests to extend the map p to all of Cn \ {0} by defining

P : Cn \ {0} −→ CPn by P (z0, . . . , zn) := [z0, . . . , zn].

Then p∗ is the restriction of P∗ to TxS2n+1 ⊂ Tx(Cn \ {0}) = Cn+1.

To calculate P∗ we use the standard charts

CPn ⊃ Ui = {[z0, . . . , zn] ∈ CPn} ϕi−→ U ′i = Cn

[z0, . . . , zn] 7→ (z0

zi, . . . ,

zizi, . . . ,

znzi

)

to identify the tangent space Tp(z)CPn at a point p(z) ∈ Ui with Tϕi(p(z))U′i = Cn via the

differential (ϕi)∗ of the local diffeomorphism ϕi.To calculate (ϕi)∗ ◦ P∗ = (ϕi ◦ P )∗ we first compute

ϕi ◦ P (z0, . . . , zn) = ϕi([z0, . . . , zn]) = (z0

zi, . . . ,

zizi, . . . ,

znzi

). (9.1)

Then the differential of ϕi ◦ P at z = (z0, . . . , zn) ∈ Cn+1 with zi 6= 0 can be calculated as

D(ϕi ◦ P )z(w) =∂ϕ ◦ P (z + sw)

∂s |s=0.

By equation (??) the k-th component of (ϕi ◦ P )(z) ∈ Cn has the form zj/zi for j = k − 1for k < i and j = k for k ≥ i. Hence the k-th component of D(ϕi ◦ P )z(w) is given by

∂

∂s |s=0

zj + swjzi + swi

=wjzi − zjwi

z2i

. (9.2)

We claim that the differential D(ϕi ◦ P )z : Tz(Cn+1 \ {0})→ TP (z)Cn = Cn is surjective. Toshow that a vector v = (v1, . . . , vn) ∈ Cn is in the image of D(ϕi ◦ P )z, we need to find avector w = (w0, . . . , wn) ∈ Cn+1 such that for each k = 1, . . . , n we have

wjzi − zjwiz2i

= vk, (9.3)

44

where as above j = k− 1 for k < i and j = k for k ≥ i. This is easy to solve: choose wi = 0;since zi 6= 0 by assumption, we can solve equation (9.3) uniquely for wj. This shows that thedifferential of P : Cn \ {0} → CPn is surjective. To show that the differential of p (which isthe restriction of P to S2n+1 ⊂ Cn \ {0}) is surjective, we need to argue that we can choosethe vector w ∈ Cn+1 to be perpendicular to z ∈ Cn+1. We observe that if w = λz for someλ ∈ C, then the calculation (9.2) shows that w is in the kernel of DPz. In particular, DPz(w)doesn’t change if we replace w by its projection to z⊥.

5. Let h : RPn → R be the function defined by

h([x0, . . . , xn]) = (n∑`=0

`x2`)(x

20 + x2

1 + · · ·+ x2n)−1.

(a) Show that h is a smooth function.

(b) Determine the critical points of h, i.e., the points of RPn which are not a regular pointsfor h.

Hint: Use the smooth atlas consisting of the charts RPn ⊃ Uiϕi−→ Dn with

Ui = {[x0, . . . , xn] ∈ RPn | xi 6= 0} and ϕ−1(v1, . . . , vn) = [v1, . . . , vi,√

1− ||v||2, vi+1, . . . , vn].

Proof. Part (a). To show that h is smooth it suffices to show that the composition

h ◦ ϕ−1i : Dn → R

is smooth for 0 ≤ i ≤ n. We compute:

(h ◦ ϕ−1i )(v1, . . . , vn) = h([v1, . . . , vi,

√1− ||v||2, vi+1, . . . , vn])

=i−1∑`=0

`v2`+1 + i(1− ||v||2) +

n∑`=i+1

`v2`

=i−1∑`=0

(`− i)v2`+1 + i+

n∑`=i+1

(`− i)v2`

(9.4)

This shows that h◦ϕ−1i is a quadratic polynomial and hence in particular a smooth function.

Part (b). A point [x0, . . . , xn] ∈ Ui is a critical points of h if and only if (v1, . . . , vn) =ϕ([x0, . . . , xn]) ∈ Dn is a critical point of f := h ◦ ϕ−1

i . Moreover, v = (v1, . . . , vn) is acritical point of f if and only if the gradient of f vanishes at v. We compute:

grad f = (∂f

∂v1

, . . . ,∂f

∂vn) = (2(−i)v1, 2(1−i)v2, . . . , 2(i−1−i)vi, 2(i+1−i)vi+1, . . . , 2(n−1)vn).

Since the coefficient in front of vk is non-zero for each 1 ≤ k ≤ n, this shows that the gradientsof f only vanishes at (v1, . . . , vn) = (0, . . . , 0). This implies that each open subset Ui ⊂ RPn,0 ≤ i ≤ n contains exactly one critical point, namely φ−1

i (0, . . . , 0) = [0, . . . , 0, 1, 0, . . . , 0](the non-zero component is in the i-th slot).

45


1. Show that the bracket of vector fields satisfies the Jacobi identity, that is, if U, V,W arevector fields on a smooth manifold M , then

[[U, V ],W ] + [[W,U ], V ] + [[V,W ], U ] = 0.

In particular, the bracket of vector fields gives the vector space X(M) of vector fields on Mthe structure of a Lie algebra (the other two requirements for the bracket, its linearity ineach slot and its skew symmetry are evident).

Proof. We recall that [U, V ] = UV − V U , where UV resp. V U are the compositions asendomorphisms of the vector space C∞(M). We compute:

[[U, V ],W ] = [UV−V U,W ] = (UV−V U)W−W (UV−V U) = UVW−V UW−WUV+WV U.

Cyclically permuting the symbols U , V , W then leads to the following 12 term formula:

[[U, V ],W ] + [[W,U ], V ] + [[V,W ], U ] =UVW − V UW −WUV +WV U

+VWU −WV U − UVW + UWV

+WUV − UWV − VWU + V UW

Inspection shows that each of the six possible permutations of U , V , W occurs twice withopposite sign, thus proving the Jacobi identity for the Lie bracket of vector fields.

2. Let V x, V y, V z be the following vector fields in R3 which are the infinitesimal generatorsof rotation around the x-axis (resp. y-axis resp. z-axis).

V x = y∂

∂z− z ∂

∂yV y = z

∂

∂x− x ∂

∂zV z = x

∂

∂y− y ∂

∂x.

Show that[V x, V y] = −V z [V y, V z] = −V x [V z, V x] = −V y.

In particular, the span of these vector fields in the space X(R3) of vector fields on R3 is aLie algebra of dimension 3.

Proof. For f ∈ C∞(R3) we have

[VxVy]f = VxVyf − VyVxf.

Let us first calculate VxVyf :

VxVyf =

(y∂

∂z− z ∂

∂y

)(z∂f

∂x− x∂f

∂z

)=y

∂f

∂x+ yz

∂2f

∂z∂x− z2 ∂

2f

∂y∂x− yx∂

2f

∂z2+ zx

∂2f

∂y∂z

46

We note that there are five terms, since evaluating the vector field y ∂∂z

on the product

function z ∂f∂x

necessitates the use of the product rule. The corresponding expression forVyVxf is obtained by simply permuting the symbols x, y on the right hand side of theprevious equation. We see that for their difference VxVyf−VyVxf most terms cancel, leavingus only with

VxVyf − VyVxf = y∂f

∂x− x∂f

∂y= −Vz.

The other two equations follow by cyclically permuting the symbols x, y, z.

3. Let F : M → N be a smooth map, and let F ∗ : C∞(N) → C∞(M) be the induced mapof functions given by f 7→ f ◦ F .

(a) Let x ∈ M and let v ∈ TxM = Derx(C∞(M),R) be a tangent vector. Show that the

composition

C∞(N) F ∗ // C∞(M) v // R

is an element of TF (x)N = DerF (x)(C∞(N),R). We remark that this shows that we can

define the differential F∗ : TxM → TyN by v 7→ v ◦ F ∗.

(b) Let V ∈ X(M) = Der(C∞(M)) be a vector field. For x ∈ M define Vx to be thecomposition

C∞(M) V // C∞(M)evx // R

where evx is the evaluation map which sends f ∈ C∞(M) to f(x) ∈ R. Show that Vx isan element of Derx(C

∞(M),R). In other words, Vx is a tangent vector at the point x.

(c) Let V ∈ X(M), W ∈ X(N) be vector fields on M resp. N . Show that the following twoconditions are equivalent.

(i) F∗(Vx) = WF (x) ∈ TF (x)N for every point x ∈M .

(ii) V ◦ F ∗ = F ∗ ◦W : C∞(N)→ C∞(M).

If these conditions are met, then the vector fields V,W are called F -related.

(d) Show that the Lie bracket is natural in the following sense. Let Vi ∈ X(M) be vectorfields that are F -related to vector fields Wi ∈ X(N) for i = 1, 2. Show that [V1, V2] isF -related to [W1,W2].

Proof. Part (a). We observe that F ∗ : C∞(N)→ C∞(M) is an algebra homomomorphism,since for f, g ∈ C∞(N), x ∈M we have

(F ∗(fg))(x) = (fg)(F (x)) = f(F (x))g(F (x)) = ((F ∗f)(x))((F ∗g)(x)) = ((F ∗f)(F ∗g))(x).

It follows that v ◦ F ∗ is a derivation at F (x) since

(v ◦ F ∗)(fg) =v(F ∗(fg)) = v(F ∗(f)F ∗(g))

=v(F ∗(f))(F ∗(g)(x)) + (F ∗(f)(x))v(F ∗(g))

=(v ◦ F ∗)(f)g(F (x)) + f(F (x))(v ◦ F ∗)(g).

47

Part (b). For f, g ∈ C∞(M) we have

Vx(fg) = evx(V (fg)) = evx(V (f)g + fV (g))

=(V (f)(x))g(x) + f(x)(V (g)(x)

=(evx(V (f)))g(x) + f(x)(evx(V (g)))

which proves that evx ◦V is a derivation at x ∈M .

Part (c). Evaluating the tangent vector of both sides of the equation in part (i) on afunction f ∈ C∞(N) we obtain:

(F∗(Vx))(f) = Vx(F∗f) = evx(V (F ∗f)) = ((V ◦ F ∗)(f))(x)

WF (x)(f) = evF (x)(W (f)) = (W (f))(F (x)) = (F ∗(W (f)))(x) = ((F ∗ ◦W )(f))(x)

This shows that evaluating the left (resp. right) hand side of equation (i) on a functionf ∈ C∞(N) is equal to the left (resp. right) hand side of equation (ii) evaluated on f (givingelements of C∞(M)), and then evaluated at the point x ∈M of part (i), giving real numbers.In particular, the conditions (i) and (ii) are equivalent.

Part (d).

[V1, V2]F ∗ = V1V2F∗ − V2V1F

∗ = V1F∗W2 − V2F

∗W1 = F ∗W1W2 − F ∗W2W1 = F ∗[W1,W2]

Here the second and third equation hold since ViF∗ = F ∗Wi, the assumption that the vector

field Vi is F -related to Wi. This shows that [V1, V2] is F -related to [W1,W2].

4. Let G be a Lie group. For g ∈ G let Lg : G → G, h 7→ gh be the smooth map given byleft multiplication by g.

(i) Show that the following conditions are equivalent for a vector field V ∈ X(G):

(a) (Lg)∗(Vh) = Vgh ∈ TghG for all g, h ∈ G;

(b) L∗g ◦ V = V ◦ L∗g for all g ∈ G.

A vector field on G satisfying these conditions is called left invariant.

(ii) Show that if V , W are left invariant vector fields on G, then so is their bracket [V,W ].In particular, the vector space of left-invariant vector fields on G is a Lie algebra, calledthe Lie algebra of G.

Proof. Part (a). These conditions are equivalent by part (c) of the previous problem.

Part (b). Comparing the definition of left-invariance with the definition of F -related vectorfields of part (c) of the previous problem, we see that a vector field V on G is left invariantif and only if V is Lg-related to itself for all g ∈ G. Part (d) of the previous problem showsthat if V and W are Lg-related to V (resp. W ), then [V,W ] is Lg-related to [V,W ] whichproves the statement.

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