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Introduction to Algebraic andGeometric Topology
Week 10
Domingo Toledo
University of Utah
Fall 2017
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Recall: Quotient TopologyI Definition
Let1. (X , TX ) be a topological space,2. Y a set3. f : X ! Y a surjective map.
The quotient topology TY on Y , is defined as
TY = {U ⇢ Y | f�1(U) 2 TX}
I TY is the largest topology on Y that makes f
continuous.
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Recall: IdentificationI Definition
Let1. f : (X , TX ), (Y , TY ) be topological spaces.2. f : X ! Y a surjective map.
f is called an identification if and only if TY is the quotienttopology just defined:
TY = {U ⇢ Y | f�1(U) 2 TX}
I Equivalent statements: A surjective map f : X ! Y isan identification if and only if
I
U open in Y () f�1(U) is open in X
I
F closed in Y () f�1(F ) is closed in X
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Recall: Examples
1. f1 : R ! S1 defined by f (t) = (cos t , sin t).
2. f2 : [0, 2⇡) ! S1 same formula.
3. f3 : [0, 2⇡] ! S1 same formula.
1 and 3 are identifications, 2 is not.
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Sufficient Conditions for Identification1. Recall definition ot open map and closed map
2. f : X ! Y continuous, surjective and open =)identification.
3. f : X ! Y continuous, surjective and closed =)identification.
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Checking Identifications
I Useful facts:I Suppose f : X ! Y , A ⇢ X ,B ⇢ Y . Then
1. f (f�1(B)) ⇢ B
2. If f is surjective, f (f�1(B)) = B
3. A ⇢ f�1(f (A)).
4. If f is surjective, then
A = f�1(B) for some B ⇢ Y () A = f
�1(f (A))
and, in this case, B = f (A).
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Continuous Maps
I X ,Y ,Z topological spaces.I f : X ! Y identification,I g : Y ! Z a map.I Then g is continuous () g � f is continuous
Xg�f�! Z
f
??y % g
Y
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I Proof:I If U ⇢ Z , then
(g � f )�1(U) = f�1(g�1(U))
I Thus, if U ⇢ Z ,
g�1(U) is open () (g � f )�1(U) is open
I Thus g continuous () g � f is continuous.
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I Equivalent Formulation:I X ,Y ,Z topological spaces, f : X ! Y an
identification.I h : X ! Z a map that is constant on the fibers f�1(y)
of f .I Then the map g in the following diagram is defined:
Xh�! Z
f
??y % g
Y
I g is continuous () h is continuous.
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I Example: Periodic functions h : R ! R
R h�! Rf
??y % g
S1
where f (t) = (cos t , sin t).
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I Example: Doubly periodic functions h : R2 ! Rh(s+2⇡, t) = h(s, t) and h(s, t+2⇡) = h(s, t) 8(s, t) 2 R2
I Let T be the torus S1 ⇥ S1
T = {(x1, x2, x3, x4) 2 R4 | x21 + x
22 = x
23 + x
24 = 1}
and f : R2 ! T be defined by
f (s, t) = (cos s, sin s, cos t , sin t).
I Then f is an identification and h is continuous () g
is continuous:
R2 h�! Rf
??y % g
T
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Equivalence RelationsI f : X ! Y surjective map of sets ()
equivalence relation on X :
x1 ⇠ x2 () f (x1) = f (x2).
I f : X ! Y surjective map of sets ()
Partition of X into disjoint subsets
X =a
y2Y
f�1(y)
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Connected Components
I Let X be a topological space. Define a relation
x ⇠ y () 9 a connected C ⇢ X with x , y 2 C
I TheoremThe relation just defined is an equivalence relation.
I Proof.Clearly x ⇠ x and x ⇠ y () y ⇠ x .Transitivity x ⇠ y and y ⇠ z =) x ⇠ z follows from thenext lemma.
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LemmaIf C1,C2 ⇢ X are connected and C1 \ C2 6= ;, then
C1 [ C2 is connected.
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Better Lemma:
Lemma
I Let {C↵}↵2A be a collection of connected subsets of
X indexed by a set A. Suppose that \↵C↵ 6= ;, Then
[↵C↵ is connected.
I Suppose C ⇢ X is connected. Then its closure C is
connected.
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I DefinitionThe equivalence classes of the equivalence relation justdefined are called the connected components of X .
I If x 2 X , let Cx be the connected component of X
containing x .
TheoremI Cx is the largest connected subset of X containing x.
I Cx is closed in X.
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DefinitionLet ⇡0(X ) denote the set of connected components of X .
I There is a surjective map ⇡ : X ! ⇡0(X ) defined by
⇡(x) = Cx ,
the connected component of X containing x .I card(⇡0(X )) = 1 () X is connected.I In general, card(⇡0(X )) is the number of connected
components of X .I ⇡0(X ) is a topological space, with the quotient
topology.
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I If f : X ! Y is continuous, there is a map, first of sets,
f⇤ : ⇡0(X ) ! ⇡0(Y )
defined byf⇤(C
X
x) ! C
Y
f (x)
where CX
xand CY
ydenote the connected components
of x in X and of y in Y respectively.
I Exercise 1: f⇤ is well-defined:x ⇠ x 0 =) CY
f (x) = CY
f (x 0).
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I Reason: x ⇠ x 0 () CX
x= CX
x 0
and f (CX
x) is a connected subset of Y contaning f (x),
hencef (CX
x) ⇢ CY
f (x).
I Exercise 2: f homeomorphism =) f⇤ bijective.
I Exercise 3: f⇤ is continuous.
I Exercise 4: f homeomorphism =) f⇤homeomorphism.
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Examples
I ⇡0(Rn) has cardinality one for n = 1, 2, . . . .
I If X is discrete ⇡ : X ! ⇡0(X ) is a homeomorphism.
I Homework Problem : X locally connected =) ⇡0(X )is discrete.
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TheoremFor n � 2, R is not homeomorphic to Rn.
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I Same argument: the unit interval I is nothomeomorphic to I ⇥ I.
I This was used in the proof that the Euclidean andtaxi-cab metrics not being isometric.
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TheoremLet X be the subspace of R2 of the letter ”X”. Let x0 be
the junction point of the four ”branches” of X , and let
x1, x2, x3, x4 be the other endpoints of the branches. Let
f : X ! X be a homeomorphism.
Then f (x0) = x0 and f permutes the points x1, . . . , x4.
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I If X is the Cantor set, what is ⇡0(X )?I Some information in the homework.
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I Let’s go back to the definition of f⇤:
I If f : X ! Y is continuous, there is a map, first of sets,
f⇤ : ⇡0(X ) ! ⇡0(Y )
defined byf⇤(C
X
x) ! C
Y
f (x)
where CX
xand CY
ydenote the connected components
of x in X and of y in Y respectively.
I Exercise 1: f⇤ is well-defined:x ⇠ x 0 =) CY
f (x) = CY
f (x 0).
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I Better to first say:
Theorem1. The connected components of X are the maximal
connected subsets of X .
2. Every non-empty connected subset of X is contained
in a unique maximal connected subset.
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I Then CY
f (x) is the maximal connected subspace of Y
containing f (Cx).I So is CY
f (x 0) for any x 0 2 Cx .I So CY
f (x) = CY
f (x 0)
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I In the same spirit as looking at homeomophisms ofthe letter X :
I Exercise: Classify the (capital) letters A,B,C, . . .Z ofthe Roman alphabet up to homeomorphism.
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Surfaces
DefinitionA topological surface is a Hausdorff space with acountable basis that is locally homeomorphic to R2.
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Examples
I An open subset of R2
I The unit sphere S2 ⇢ R3
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I Let u ⇢ R3 be open, let f : U ! R be a function ofclass C1, and suppose that the gradient rx f 6= 0 atany point x where f (x) = 0. Then
{x 2 R3 |f (x) = 0}
is a topological surface.I Reason: The implicit function theorem.
(Will review and prove)
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Surfaces as Identification Spaces
Example: Torus
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Surface of Genus Two
Figure: Identifying Octagon
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Figure: Surface of Genus Two
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Non-Hausdorff Quotient Spaces
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A non-Hausdorff “Surface”
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Existence theorems based on connectedness
I The intermediate value theorem.
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I The implicit function theorem.
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