solutions to some of the exercises from tennison’s sheaf theory · 2011-06-19 · 2 exercises in...

Solutions to some of the exercises fromTennison’s Sheaf Theory

Pieter Belmans

June 19, 2011

Contents

1 Exercises at the end of Chapter 1 1

2 Exercises in Chapter 2 6


4 Exercises in Chapter 3 12



1 Exercises at the end of Chapter 1

1. Prove that the lim−→ of the direct systems of the examples is F(;) in each case.Generalise.

Solution The direct system of a topology corresponds to a complete lattice,hence there is a minimum and a maximum: ; and X . The direct limit nowcorresponds to ‘the object that goes to the right of everything’ but in thiscase, ; is on the right over everything, therefore, the direct limit is givenby F(;).

This holds for every direct limit: if there is a single object in the direct systemthat goes to the right of entire system it corresponds to the limit.

1

2. Prove directly from the definitions that if U is a direct limit of a directsystem (Uα)α∈Λ of sets, then U =

⋃

α∈Λ Im(Uα→ U).

Solution The ρα,β in the definition of a target are injective maps. Nowthe set

∐

α∈Λ Uα is a possible target, with σα mapping an element to allpositions in the disjoint union such that it is contained in the correspondingset Uγ.

By the definition of the direct limit, there must be a unique f to this target,making the triangle commute.

3. a) Interpret and prove: a set is the direct limit of its finite subsets.

Solution The construction of the direct limit consists of quotientingthe disjoint union. All elements of the set U are represented in a finitesubset (for instance the singletons), so the disjoint union containsall elements. Elements are now identified if they are identified in thepower set lattice of U , but as the union of two finite subsets is still finitethe identification of an element v ∈ Uα, Uβ occurs in Uγ = Uα ∪ Uβ .

b) Interpret and prove: an abelian group is the direct limit of its finitelygenerated subgroups.

Solution Analogously, finitely generated subgroups are representedin the same kind of lattice structure with atoms and finite unions ofgenerators.

c) Can you obtain Z as a direct limit of finite abelian groups?

Solution No, only torsion groups are obtainable. As finite abeliangroups are all isomorphic to direct sums of Z/pkZ and we’ll be takinga quotient from a direct sum, all elements will keep their finite order.

4. a) Characterise direct systems of sets with lim−→= ;.Solution As the direct limit consists of the disjoint union modulo somerelations, the direct limit is always nonempty, unless the disjoint unionitself consists of empty sets. Therefore all sets in the direct system mustbe empty: you cannot obtain an empty quotient from a nonempty set.

b) Produce an interesting direct system of abelian groups with lim−→= 0,the trivial group. Characterise such systems.

Solution As the construction of the direct limit consists of taking thedirect sum of all groups and taking the quotient with the subgroupgenerated by all iα(gα)− iβ(resα,β(gα)) we need to obtain all elementsof the direct sum. Hence iβ(resα,β(gα))must be zero for all (α,β) ∈ Λ1.So resα,β must kill all elements, we therefore have rather boring re-striction morphisms.

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5. What can you say about the direct limit of a direct system all of whose mapsare injective? Surjective?

Solution All the maps to the direct limit are injective (or surjective). Inthe injective case the direct limit contains information on all objects of thedirect system, in the surjective case the direct limit contains the commoninformation of all objects.

6. For n ∈ N0, let Cn(x) denote a cyclic group of order n with generator x .Let p ∈ N be a prime number. Let G be the direct limit of the following directsystem of abelian groups:

0= Cp0(x0)→ Cp1(x1)→ . . .→ Cpn(xn)→ Cpn+1(xn+1)→ . . . (1)

(where Cpn(xn)→ Cpn+1(xn+1) takes xn 7→ pxn+1). Preferably without re-sorting to the explicit construction prove:

a) G is infinite, but torsion (i.e. element has finite order).

Solution Based on the previous exercise we realize that G is a bigstructure as every map is injective. As every step in the chain adds afinite number of elements and there are countable steps, we obtain acountable infinite group.

As the construction of the direct limit consists of summing all objects ofthe direct system and taking a quotient, we get that every element ofthe direct sum has finite order (namely the order of the highest term)and the quotient keeps this structure.

b) Every finitely-generated subgroup of G is finite. Find all of them.

Solution Every finite set of generators has a maximal index n suchthat the term for Cpn is nonzero but for Cpn+1 is zero. The order of thefinitely-generated subgroup is now limited by the order pn.

Deduce that G has no proper infinite subgroup, and no maximal propersubgroup. Can either of these situations arise for subspaces of a vector space(using dimension instead of order)? Identify a realisation of G inside theunit circle S1 ⊆ C (under multiplication).

Solution The construction of the direct sum learns us that there must be aninfinite number of nonzero summands, but as every Cpn is trivially embeddedin Cpm for all m> n and every such instance is identified we cannot get to aproper infinite subgroup. Analogously there is no maximal proper subgroup.

The only interesting case occurs in infinite-dimensional vector spaces, thefinite-dimensional case is trivial (take the quotient of the space with thefield).

The obtained structure is the Prüfer group.

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7. Consider the following direct system of abelian groups: fix r ∈ Z; forall n ∈ N let Un = Z and for n ≥ m let ρm,n : Um → Un be multiplicationby rn−m. Identify the lim−→ as a subring of Q.

Solution This is the localization at the ideal generated by r. In the caseof r prime this is Z(p), in the other case we obtain all fractions with primefactors of r as denominators.

8. Interpret and prove: the direct limit of a system of exact sequences is exact.

Solution For 0 → An → Bn → Cn → 0 an exact sequence all the mapsdescribed in the definition of a direct limit are commuting by the propertiesof injective and surjective maps.

9. The notions of target and direct limit can be formulated without the re-striction (a) of Definitions 3.1 and 3.11. What difference does this maketo the constructions? Find a system of abelian groups (in this generalisedsense) with direct limit A⊕ B without having this abelian group appear inthe system. Justify Remark 3.21.

Solution Just taking a quotient won’t work, there are several componentsin the generalised direct system that must be handled appropriately.

Take X :=]0,1[∪]1,2[ and the Euclidean topology with X removed fromthe topology. Now the direct system doesn’t satisfy the conditions, but bytaking a direct system on ]0,1[ that produces A as its direct limit (in theformer sense) and likewise on ]1, 2[ one that produces B we obtain a systemof abelian groups in the generalised sense that produces A⊕ B.

10. Formulate the dual notions of inverse system and inverse limit lim←− (reversethe arrows).

Solution As the exercise suggests, this is just a reversal of the arrows ρα,βto Uβ → Uα, for α≤ β .

Find inverse systems:

a) of finite sets whose lim←− is infinite;

Solution Take (N,≤) and define Ak = 0, . . . , k. We get ρi, j : A j → Aithe projection for i ≤ j. The direct limit is the set N.

b) of finite abelian groups whose lim←− is infinite;

Solution The ring of p-adic integers Zp is given by lim←−Z/pnZ.

c) of abelian groups whose lim←− is Z (without Z in the system).

Solution Start from Q. Let pn denote the nth prime number. Nowdefine Gk to be the subgroup of Q in which all primes but p1, . . . , pnare allowed as factors of the denumerator. The morphisms are theinjections. Now Z is the inverse limit, it being the subgroup of Q inwhich all prime factors of denumerators are removed.

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11. Verify that if (Rα)α∈Λ is a direct system of abelian groups such that each Rαis a ring and all the ρα,β are ring morphisms, then lim−→Rα has a natural ringstructure such that all the maps Rβ → lim−→Rα are ring morphisms.

Solution The ring structure is preserved by the direct sum construction ofthe direct limit of abelian groups. As the subgroup H1 is the construction isan ideal because the multiplicative structure is preserved by the injections,we obtain a quotient ring.

12. What are the stalks of the presheaf P2 of 2.E?

Solution We have (P2)x0= Z as every open subset U containing x0 will

have P2(U) = Z, with ρα,β = idZ in the direct system.

For (P2)x with x 6= x0 we get 0, the trivial group. We have |x− x0| = ε > 0,so for the open set B(x ,ε)∩ [0,1] the presheaf gives 0, all values of thepresheaf from there on are the same trivial group.

13. Construct a topological space X and a presheaf F of abelian groups on Xwith the properties:

a) for any open U ⊆ X : F(U) 6= 0;

b) for all x ∈ X the stalk Fx = 0.

If you cannot, prove that it is impossible. Compare with 4(b).

Solution This is impossible, exactly by Q4(b). We’d need restriction mapsrestricting everything to zero, but by the sheaf conditions this is impossible.

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2 Exercises in Chapter 2

Exercise (1.9). Show that if G is an abelian sheaf, then G(;) = 0, the trivialgroup.

Solution By taking the empty cover of the empty set, the first part of the sequencethat corresponds to the equalizer diagram reduces to

0→ G(;)→ 0 (2)

because we’re taking the product over an empty set (Γ = ;). Now G(;) is embed-ded in 0, and must therefore be equal to the trivial group.

Exercise (1.12). Find an example with G not a monopresheaf where 1.10 fails.

Solution Take both F and G the presheaf P1 from Example 1.2.E. Now de-fine f : F → G to be the trivial map for F(U) → G(U), U 6= X open andlet f (X ): F(X ) → G(X ) : n 7→ 2n. Similarly, g(X ): F(X ) → G(X ) : n 7→ 3n.Now f and g agree on all stalks but as the monopresheaf fails (all restrictions areelements of the trivial group but the global behaviour differs) we have f 6= g.

Exercise (3.7). Check the functorial properties¨

Γ ( f g) = Γ f Γ g

Γ (id) = id(3)

Solution We have a morphism of sheaf spaces id: E→ E, obtaining a morphismof sections Γ (U , E) → Γ (U , E) : σ 7→ idσ, which is locally an identity. Thisextends to the entire sheaf Γ (E), so Γ (id) = id.

As Γ (g f ) is defined locally as a morphism of sections

Γ (U , E)→ Γ (U , E′′) : σ 7→ g f σ (4)

in which f σ ∈ Γ (U , E′) by definition of Γ ( f ), we obtained the desired property.

Check also that if f : E→ E′ is a morphism of sheaf spaces over X , then

Γ f

x : (Γ E)x →

Γ E′

x (5)

and

f |p−1(x) : p−1(x)→ p′−1(x) (6)

are isomorphic maps.

Solution This is nothing but Proposition 3.6.

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Exercise (3.9). Check the functorial properties

¨

L( f g) = L f Lg

L(id) = id(7)

Solution The map idx : Fx → Fx is applied for all x and L(F) is the disjoint unionof these stalks, so L(id) = id.

The composition is handled as in Γ ( f g).

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1. Let I = [0, 1] ,→ R. Show that there is a unique (up to isomorphism) sheaf Fon I with stalks: F0 = F1 = Z, Fx = 0 if x ∈ I \ 0, 1.

Solution A presheaf has a unique sheafification, so we need to define apresheaf with these stalks and sheafify it. Do this by letting Γ (U , E) = Zif 0, 1 ∩ U 6= ; and Γ (U , E) = 0 otherwise. Now the stalks agreeand Γ (E) = F .

What is Γ (I , F)?

Solution It is Z⊕ Z, as it is composed of contributions of the stalks in 0and 1.

Let G be the constant sheaf Z on I . How many morphisms are there from Fto G? From G to F?

Solution To make the triangle in the definition of a (pre)sheaf morphismcommute, observe that resU

V is either the identity1 of the projection on 0while res′UV is the identity for all open sets. So for a morphism from F to Gwe have on option: the identity on U containing 0 or 1 and the canonicalinjection on the other open sets.

For a morphism from G to F on the other hand we have total freedom for Unot containing 0 or 1: whatever we do on these open sets doesn’t matter asthe restriction morphism kills it.

2. Show that the following conditions are equivalent for a topological space X :

a) X is locally connected (that is, each point has a base of connectedneighbourhoods);

b) for any set A, the constant sheaf AX has Γ(U , AX ) =∏

t∈U ′ A for U openin A where U ′ is the set of connected components of U;

c) 2b holds for A= 0,1, some set with two elements.

Solution

2a⇒ 2b We have that

Γ (U , AX ) =

locally constant functions U → A

(8)

and as any covering of U falls apart in disjoint coverings of everyconnected component of U we obtain

∏

t∈U ′ A.

2b⇒ 2c Trivial specialization.

1Or some isomorphic map, the image of 1 defines the entire morphism and every morphismcorresponds to a multiplication with a scalar.

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2c⇒ 2a As Γ (U , 0,1X ) =∏

t∈U ′ 0,1 we obtain a notion of indicatorfunctions. They determine uniquely to which connected componentsan open subset of U belongs.

When these conditions hold, what are the restriction maps in terms of therepresentation given in 2b?

Solution The restrictions are projections on the remaining components.For V ⊆ U open we have the inclusion V ′ ⊆ U ′ of sets of connected compo-nents.

3. Let F be a presheaf on a space X , and let V be open in X . Then we candefine a presheaf F |V on V by the same recipe as F ; that is (F |V )(U) = F(U)for U open in V .

Show that is F is a sheaf, so is F |V .

Solution Intersecting an open covering with an open set V leads to an opencovering of this set in V in the topology on X . Now the conditions for thesheaf on V trivially hold because they hold in X .

Show also that if F has sheaf space p : L(F)→ X , then F |V has sheaf space(p−1(V ), p|p−1(V )). What can you say when V is not open?

Solution The projection map p must be restricted to p−1(V ) as the cor-responding sheaf acts on V . Now Lemma 3.5(b) provides the necessaryconditions, as intersections of open sets are open.

4. Let F be a sheaf on a space X with sheaf space L(F)p→ X , and let A be

a subspace of X . We ca define the set (or abelian group) of sections of Fover A by

Γ (A, F) = Γ (A, LF) (9)

=n

sections of the continuous map p−1(A)p→ A

o

. (10)

Show that we can define Γ (A, F) in terms of F alone as Γ (A, F) = lim−→Γ (U , F)where the direct limit is taken over the set of open subsets U of X suchthat U ⊇ A. (Colloquially, this says that a section of F over A extends uniquelyinto a small neighbourhood of A.)

Solution A section over A is a section for p : p−1(A)→ A, but continuity isdefined in terms of open sets, so every section over A must be a section forsome U ⊇ A. By taking the direct limit we identify equal sections.

5. Let F be a sheaf on a space X and let (Mi)i∈I be a locally finite covering of Xby closed sets (so that for each x ∈ X ,

i ∈ I | x ∈ Mi

is finite). In the nota-tion of Q4, suppose we are given a family (si)i∈I with ∀i ∈ I : si ∈ Γ (Mi , F)and ∀i, j ∈ I : si = s j on Mi ∩M j .

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Show that there is a unique s ∈ Γ (X , F) with ∀i ∈ I : s = si on Mi .

Solution By the previous exercise all sections on a closed Mi are lifteduniquely to a small (open) neighbourhood and we’ll take a finite intersectionof opens. Now we can apply the glueing condition for sheaves and find aglobal sections.

6. Let K be any infinite field and L = K(t) a simple transcendental extension.Let X be the topological space obtained by giving K the topology whoseclosed sets are the finite subsets of K .

Define a sheaf O of commutative unital rings on X as follows: for U openin X , U 6= ;, let

O (U) =§

f ∈ L

∃g, h ∈ K[t]: f =g

h∨∀P ∈ U : h(P) 6= 0

ª

. (11)

If ; 6= V ⊆ U then O (U) ⊆ O (V ) ⊆ L and we take the inclusion as therestriction map resU

V . Show that O is a sheaf of rings on X .

Solution As X is equipped with the cofinite topology, it is compact. Nowevery open covering reduces to a finite covering. Now the conditions for asheaf are fulfilled: to every open set we’ve assigned an algebra, restrictionsare correct considering them as K-valued functions and the glueing condi-tions reduces to an argument on the poles of the section (when consideredover X ).

A better answer would be to state that the sections over U are defined as theintersection of the sections over the (finite) covering of U . It is the inverselimit of algebras realized as a pullback, which is exactly this intersection forour finite covering.

Identify the stalk OP of O at P ∈ X as a subring of L and show that it is alocal ring. What is its residue field and its field of fractions?

Solution The stalk corresponds to

OP =§

f ∈ L

∃g, h ∈ K[t]: f =g

h∧ h(P) 6= 0

ª

. (12)

Now the ideal of f such that f (P) = 0 is the unique maximal ideal becausethe functions for which f (P) 6= 0 are invertible, hence not part of anymaximal ideal. The quotient gives a residue field, which is an extension fieldof K . The corresponding field of fractions is just L.

When does O have non-polynomial global sections? That is, we certainlyhave K[t]⊆ Γ (X ,O ); when is the equality strict?

Solution When K is not algebraically closed, there must be polynomi-als h ∈ K[t] such that h(P) 6= 0 for all P ∈ X . For instance t2 + 1 in R[t]provides a good denominator for non-polynomial global sections.

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This is an equivalence: a non-polynomial global section must consist of afraction of polynomials (exactly by the definition of the sections), so its de-nominator cannot have a zero. This only occurs when K is not algebraicallyclosed (otherwise the denominator factors into linear components).

We can consider f ∈ Γ (U ,O ) as a function on U , namely express f = g/has in (11) and define for P ∈ U

f (P) = g(P)/h(P) ∈ K . (13)

Show that this defines a morphism φ : O → F where F is said to be thesheaf of K-valued functions on X (do this by giving K the indiscrete topol-ogy), and that putting O ′(U) = Im

φ(U)

defines a sheaf O ′ with a mor-phism O → O ′. Prove that O → O ′ is an isomorphism of sheaves. Hence wemay regard O as a sheaf of K-valued functions on X .

Solution As g(P)/h(P) ∈ K each f defines a K-valued function on an openset U . This is a sheaf by trivial reasons of gluing and restrictions: we’ve gota pointwise definition of a function.

For the isomorphism of sheaves: for O → O ′ we obtain an inverse mapbecause functions are one-valued. For O ′→O the isomorphism is obtainedby the same argument.

As g(P)/h(P) ∈ K this easily defines a good morphism. By removing enoughpoints from an open U in order to determine enough values to uniquelydefine f on U we get a nice sheaf morphism. Now O → O ′ is a sheafmorphism because of this property.

For K = C show that O ′ is a subsheaf of the sheaf Cω of analytic C-valuedfunctions on X = C.

Solution For K = C we know that these K-valued functions corresponduniquely to rational functions, which are Cω on a suitable open set (as theyare meromorphic).

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4 Exercises in Chapter 3

Exercise (1.3). Show that identities are unique.

Solution Take B = A. If both idA and id′A fulfill the conditions of an identitymorphism, we have

idA id′A = idA by f id′A = f (14)

= id′A by idA f = f (15)

and therefore idA = id′A.

Exercise (1.8). Show that a natural transformation n is a natural equivalence ifand only if ∀A∈ Ob C we have that nA is an isomorphism.

Solution Just put the two diagrams from the definition next to eachother and

chase the arrows: F(A)nA→ G(A)

mA→ F(A) is also given by idF (A) and vice versa. Theopposite direction follows from this as well.

Exercise (3.2). If f ∈ Hom(F, G) then Ker( f ) has the universal property:

If H is a presheaf and g ∈ Hom(H, F) is such that Hg→ F

f→ G = 0

then g factors uniquely as

H

Ker( f ) F

g (16)

Solution The factorization is uniquely determined by g, as it will map the sheaf Ginto Ker( f ) in a unique (namely its own) way.

Exercise (4.2). If f ∈ Hom(F, G) then PCok( f ) has the universal property:

If H is a presheaf and g ∈ Hom(G, H) is such that Rf→ G

g→ H = 0

then g factors uniquely as

G

PCok( f ) H

g (17)

Solution Analoguous to Exercise 3.2.

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Exercise (6.2). Formulate the universal property that you would like a conceptof “image” to satisfy and verify that PIm and SIm do in the categories Presh/Xand Shv/X .

Solution Given a morphism f : F → G, we want the image of f to be the object Isuch that h: I → G is a monomorphism satisfying:

1. there exists a morphism g : F → I with f = h g;

2. any other object I ′ with morphism f ′ : X → I ′ and corresponding monomor-phism h′ : I ′→ G such that f = i′ f ′ (i.e., it tries to act like our image) wehave a unique morphism m: I → I ′ with f ′ = m g and h= h′ m.

By Theorem 4.13 we have the monomorphism and by the definition of the respec-tive cokernels we know that the images of f are mapped to zero, so the kernelof G→ PCok f or G→ SCok f is exactly the image we are looking for.

Exercise (6.3). Check that PIm( f ) is a presheaf whose abelian group of sectionsover each open U is the image of f (U), while SIm( f ) is a sheaf whose stalk ateach x ∈ X is the image of fx .

Solution By the same reasoning as Theorems 4.7 and 4.8 using Proposition 3.9.

Exercise (6.8). Prove that T is exact if and only if T preserves all exact sequences.

Solution Every general exact sequence can be reduced to a short exact sequencewhich can be embedded in the five-term sequence from the definition.

Exercise (7.4). Verify the functorial properties of φ∗ and −∗.Solution We have φ∗(idF ) = idφ∗F because by definition of the direct image of amorphism

φ∗

idF

(U) = idF

φ−1(U)

= idφ∗F (U) (18)

and id∗ = id by the definition of the direct image (pre)sheaf.

Now take Ff→ G

g→ H (pre)sheaf morphisms on continuous maps X

φ1

→ Yφ2

→ Z .We now have

φ2 φ1

∗

g f

(U) =

g f

φ2 φ1−1(U) (19)

=

g f φ1−1 φ2−1

(U) (20)

=

f φ1−1

φ2∗ g(U)

(21)

=

φ2∗ g φ1

∗ f

(U) (22)

by peeling of the definition of φ∗ f .

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1. Let P be the category of pointed sets, whose objects are the pairs (A, a)with a ∈ A∈ Ob(Sets), and whose morphisms (A, a)→ (B, b) are the mapsof sets f : A → B such that f (a) = b. Show that P is a category with azero object, kernels and cokernels, but in which not every epimorphism is acokernel.

Solution Every singleton set (x , x) is a zero object. The initial object ;from Sets is now changed to (x , x) as every morphism needs to map thebase point and hence a unique map from our initial object is defined. Theterminal object x in Sets remains unchanged as it allows for exactly onemorphism (which preserves the base point).

For kernels and cokernels we need a zero morphism, which in this caseis the projection on the base point. For a map f : (A, a)→ (B, b) we havethat Ker( f ) = ( f −1(b), a) and Coker( f ) = ((B \ f (A))∪ b , b).

The fact that not every epimorphism is a cokernel is due to split epimor-phisms being surjective, while the cokernel removes the image except forthe base point.

2. For X any topological space, show that the following presheaves of setsover X are in facts sheaves:

a) Fixing an open V ⊆ X , let

hV (U) =

(

singleton U ⊆ V

; U * V(23)

for U open in X , with the unique restrictions.Solution Both the monopresheaf and the gluing condition are mean-ingless for U * V , so they trivially hold. For U ⊆ V both conditions arevalid as there is only one possible section. Therefore all restrictionsand gluings are well-defined.

b) Let Ω(U) =

W |W ⊆ X open and W ⊆ U

for U open in X with re-striction: Ω(U)→ Ω(V ) : W 7→W ∩ V . Interpreting presheaves on Xas contravariant set-valued functors on U , the category of open setsof X , show that the hV defined by 2a are the representable functors

hV = HomU (−, V ): U 7→ HomU (U , V ). (24)

Solution First of all note that Ω(U) describes the (open) subspacetopology induced by X on U . Now we have in any topology (which isjust a lattice) that

HomU (U , V ) =

(

i : U ,→ V U ⊆ V

; U * V(25)

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which corresponds to hV = HomU (−, V ).

Interpret the Yoneda lemma as saying that

HomPresh/X (hV , F)'→ F(V ) (26)

for any presheaf F on X . Putting F = hU , this shows that V 7→ hV is a fulland faithful embedding of U into Presh/X .

Solution We have that U is locally small (as each set of homomorphismscontains either zero or one morphism, which is quite small), so by the(contravariant) Yoneda lemma we can assign to each object V ∈ U afunctor hV = Hom(−, V ) to Sets.

Now we have that for each object V ∈ U the natural transformations fromthe presheaf (of representable functor) hV to a presheaf F (which is justa functor, we are working in an functor category) correspond bijectivelyto F(V ). I.e.,

HomPresh/X (hV , F)1:1←→ F(V ) (27)

4. Let F, G be presheaves of sets on a topological space X . Define a newpresheaf Hom(F, G) with

Hom(F, G)(U) := HomSets (F(U), G(U)) (28)

for U open in X . Show that if F and G are both sheaves, then so is Hom(F, G).

Solution

Hom(F, G) presheaf By restricting a section f ∈ Hom(F, G)(U) to its re-striction resU

V ( f ) we get by commutativity in the diagram for sheafmorphisms a good section in Hom(F, G)(V ), compatible with all re-strictions, for V ⊆ U open.

Hom(F, G) monopresheaf Given the situation for the monopresheaf condi-tion, if s 6= s, both being maps F(U) → G(U) they must disagreeon some x ∈ U , i.e., s(u) 6= s(u′). But u ∈ Uλ for some λ ∈ Λ,so resU

Uλ(s)(x) 6= resU

Uλ(s′)(u), a contradiction.

Hom(F, G) glues Define the section over U by its values in the Uλ, theobvious restrictions are satisfied and we get a section glued from itscomponents.

Prove that this construction has the following universal property: for F , G, Hpresheaves (respectively sheaves) of sets on X , there is a bijection

Hom

H,Hom(F, G)∼= Hom (H × F, G) (29)

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natural in F , G and H. Here H × F is the product object provided by 3(d); itis constructed “pointwise”.

Solution Take F ∈ Hom(H,Hom(F, G)), that is for U ⊆ X open we have

Γ (U ,F ):H(U) → Hom(F, G)(U) = HomSets (F(U), G(U))

h 7→

fh : F(U)→ G(U) : x → fh(x)

.(30)

This is now mapped to Hom(H× F, G) by mapping each Γ (U ,F ) to the map

(H × F) (U) = H(U)× F(U) → G(U)(h, f ) 7→ F (h)( f ) (31)

surjectivity We can uniquely define an F ∈ Hom(H,Hom(F, G)) by thisconstruction, so the map is surjective. In more detail, define

Γ (U ,F )→H(U) → Hom(F, G)(U)

h 7→

f 7→ g (32)

and this provides a good preimage.

injectivity Assume F1 and F2 in Hom(H, Hom(G, F)) are mapped to thesame element in Hom(H × F, G). Then by the construction above theinverse image of this element corresponds to a unique element in thedomain, a contradiction.

Show that, if we want the property (29) to hold, then the definition of Homis forced on us, for presheaves at least.

Solution Taking H = hU , now we want

Hom(hU ,Hom(F, G))∼= Hom(hU × F, G) (33)

to hold. But on all V ⊆ U this reduces to HomSets(F(U), G(U)) becausein Sets taking a product with a singleton is an isomorphism.

Reinterpret (29) as requiring the existence of an evaluation map

Hom(F, G)× F → G (34)

with a suitable universal property.

5. Show that each of the categories Presh/X and Shv/X of Q3 has a subobjectclassifier, that is, an object Ω with the equivalent properties (prove theirequivalence):

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a) there is a natural bijection

Hom(F,Ω)∼= subobjects of F (35)

for any object F ;

b) Ω has a special subobject 1t→ Ω such that any morphism G→ F is the

pullback of t over a unique morphism F → Ω (called the classifyingmap of G→ F).

Proof. As 1→ Ω is a monomorphism, we have that the realization of thepullback over F → Ω is again a monomorphism. As pullbacks are unique (upto isomorphism) we find the bijection (and vice versa every map g : F → Ωprovides a pullback scenario).

Solution

Presh/X By statement 5a Hom(F,Ω) bijects with the set of subobjects of F ,which is a set of objects in Presh/X . Now take F = hU , by the Yonedalemma we have

Hom(hU ,Ω) = Ω(U)1:1←→

subobjects of hU

. (36)

But a subobject of hU is an hV for V ⊆ U open, so by the full andfaithful embedding of U into Presh/X , we have

Ω(U) =

V ⊆ X |V open, V ⊆ U

. (37)

For a general presheaf F we now find

Ω(U) = P (F(U)) . (38)

Shv/X We want to associate a map f : F → Ω to every subsheaf G. Define

f (U): F(U)→ Ω(U) : x 7→⋃¦

V ⊆ X open |ρUV (x) ∈ G(V )

©

. (39)

This defines a bijection between morphisms from F to Ω and subobjectsof F as the map exactly describes how a subobject is constructed, givingfor each section over U how much it is a subobject.

Compare the situation in Sets = Shv/X for X a one-point space, where Ω isa two-point set and the classifying map is the characteristic map of a subset.

Solution Now t : 0 → 0, 1, the classifying map is the characteristicfunction and G is a subobject of F if and only if it is a subset of F(X ) ∼= F .Saying the classifying map is the characteristic function is odd, as thebijection found concerns the characteristic functions from F to 0,1, hencethey define the subsets. The pullback is just the monomorphism.

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6. Since any category of presheaves is a functor category, we can define itfor categories other than those of open sets of a topological space. For Cany category, let Presh/C be the category SetsC

opof contravariant func-

tors C → Sets (and natural transformations).

As a special case, any group G can be regarded as a category with oneobject whose endomorphisms are the elements of G, with compositiondefined as in G (hence every morphism is an isomorphism). Show that thecategory Presh/G can be regarded as the category of sets-with-a-Gop-action,that is the category of permutation representations of G.

Solution As F ∈ Presh/G is a contravariant functor G → Sets and bydefinition of the category of a group Ob(G) =

·G

, we have F(·G) a set.As all endomorphisms defined on ·G by G are invertible (G a group, notjust a monoid) we have F(End(·G)) = Aut

F(·G)

, i.e., all bijections orpermutations on F(·G). Now every g ∈ G defines an (invertible) action onthe set F(·G), in which x ∈ F(·G) is mapped to F(·G)(x).

9. Let 0 → P → Q → R → 0 be an exact sequence of presheaves of abeliangroups over a topological space. Show that

a) if Q is a sheaf and R is a monopresheaf, then P is a sheaf;

Solution As P is a subpresheaf of the sheaf Q by the exactness it is amonopresheaf using Corollary 3.7.

For the glueing condition: we can glue a section s in Q(U). Nowthe (sλ)λ are mapped to 0 in R(Uλ) and by the monopresheaf conditionwith s the image of s in R(U) and s′ = 0, we find s = 0, hence s ∈ P(U)because it is an element of the kernel of the quotient map.

b) if P is a sheaf and Q is a monopresheaf, then R is a monopresheaf.

Solution Take s ∈ R(U) and s′ = 0 as in the definition of a mono-presheaf. By the exactness we find in Q(Uλ) that the cosets of ρU

Uλ(s)

and 0 are equal by lifting the quotient. Hence ρUUλ(s)+pλ = p′λ, both pλ

and p′λ in the embedding of P(Uλ) in Q(Uλ).Now we can glue the pλ and p′λ together, from which s = 0 follows bythe monopresheaf condition on Q.

10. A sheaf F of abelian groups on X is called locally free if and only if eachpoint x ∈ X has an open neighbourhood U in X such that the sheaf F |Uis isomorphic to a constant sheaf with typical stalk a free abelian group(of finite rank). Show that if X is connected, a locally free sheaf has awell-defined rank.

Solution As it is locally free, the rank of the stalk is extended in a smallneighbourhood. But as X is connected this procedure extends to all of X .

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Show by example that a locally free sheaf on X need not be isomorphic toa constant sheaf even if X is connected, and that φ∗ does not preserve theproperty of being locally free.

Solution Take X = 0,1, 2 with topology T = 0, 1 , 1 , 1,2 , X anddefine F to be the sheaf with sections

Γ (0,1 , F) = Z (40)

Γ (1,2 , F) = Z (41)

Γ (1 , F) = Z (42)

Γ (0, 1,2 , F) = Z⊕Z (43)

with restriction morphisms the projection on the first term. Now takefor x ∈ X any open U 6= X , then F |U = UZ, but F is not isomorphic toany constant sheaf.

Prove however that the inverse image of a locally free sheaf of rank n islocally free and of the same rank.

Solution The inverse image f ∗ is given locally by the direct limit of opensets containing the inverse image of the continuous map f , but this directlimit consists of data of a locally free sheaf, hence it is locally free.


3. Draw a picture of SpecZ[t].

Solution In Figure 1 my own LATEX version of Mumford’s SpecZ[t] is given.

4. Show that for any ring R, Spec R is compact.

Solution We start from Spec R =⋃

i∈I Ui. We have for all i ∈ I an opencovering of Ui by distinguised opens, i.e., sets of the form D( f ), as thetopology is given by a basis, closed under intersection. Hence, we obtainedan open covering of Spec R using more sets such that every Ui can is exactlycovered.

Now define S the set of all these representatives of distinguised opens. Theset S cannot be contained in any proper ideal or we would find a maximalideal not yet covered, so S generates all of R. Now write unity as a finitelinear combination. The corresponding distinguised opens cover Spec R, sowe have reduced the covering to a finite covering.

9. Algebraic curves

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Topological version Show that the subset of R2 given by

C =¦

(x , y) | y2 = (x + 1)x(x − 1)©

(44)

is a manifold (topological, differentiable or analytic as you wish),whereas that given by the equation y2 = x2(x + 1) is not.

Solution In Figure 2 an impression in R2 is given. In the first caseone component is homeomorphic (or diffeomorphic as you wish) toa circle while the other is homeomorphic to R. In the second casethough, the self-intersection in the origin creates a problem: there is nocorrect definition of the tangent. Or if you are looking for a topologicalargument: removing the origin creates three components, which isimpossible for a manifold, given its local Euclidean structure.

Algebraic version Let K be any field and f (x , y) ∈ K[x , y] = R. By Q6the morphism R→ R/( f ) identifies the space Spec R/( f ) with a closedsubspace C of Spec R= A2

K = X say. Find the sheaf of ideals I in OX |Csuch that, letting O ′ be the quotient of OX |C by I , the morphism

Spec R/( f )→ (C ,O ′) (45)

is an isomorphism of ringed spaces (and so of affine schemes).

Putting f (x , y) = y2 − x3 + x or y2 − x3 − x2 we get the algebraicanalogues of the curves of part (a). What is different about the twocases?

Solution The local ring in the origin is different: it shows the oc-curence of two tangent lines. It is a singular point.

Other curves for your amusement:

y2 = x2(x − 1) (46)

y2 = x3 (47)

Solution

In Figure 3 the embedding in R2 is given. We see just one componentcontaining nothing but regular points while the curve y2 − x3 has acusp: there is a multiple tangent in the origin.

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[(2)]

V((2))

[(3)]

V((3))

[(5)]

V((5)) V((7)) . . .

[(0)] generic point

[(2, X )] [(3, X )] [(5, X )] (X )

[(2, X + 1)]

[(3, X + 1)][(5, X + 1)]

[(3, X + 2)]

[(5, X + 2)]

[(5, X + 3)]

[(5, X + 4)]

(X 2 + 1)

Figure 1: Mumford’s treasure map

x

y

(a) y2 = (x + 1)x(x − 1)

x

y

(b) y2 = x2(x + 1)

Figure 2: Plots

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x

y

(a) y2 = x2(x − 1)

x

y

(b) y2 = x3

Figure 3: Plots

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solutions to some of the exercises from tennison’s sheaf theory · 2011-06-19 · 2 exercises in...

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