Solutions to some of the exercises from Tennison s Sheaf Theory

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1 Solutions to some of the exercises from Tennison s Sheaf Theory Pieter Belmans June 19, 2011 Contents 1 Exercises at the end of Chapter Exercises in Chapter Exercises at the end of Chapter Exercises in Chapter Exercises at the end of Chapter Exercises at the end of Chapter 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 now corresponds to the object that goes to the right of everything but in this case, is on the right over everything, therefore, the direct limit is given by F(). This holds for every direct limit: if there is a single object in the direct system that goes to the right of entire system it corresponds to the limit. 1

2 2. Prove directly from the definitions that if U is a direct limit of a direct system (U α ) α Λ of sets, then U = α Λ Im(U α U). Solution The ρ α,β in the definition of a target are injective maps. Now the set α Λ U α is a possible target, with σ α mapping an element to all positions in the disjoint union such that it is contained in the corresponding set U γ. 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 quotienting the disjoint union. All elements of the set U are represented in a finite subset (for instance the singletons), so the disjoint union contains all elements. Elements are now identified if they are identified in the power set lattice of U, but as the union of two finite subsets is still finite the identification of an element v U α, U β occurs in U γ = U α U β. b) Interpret and prove: an abelian group is the direct limit of its finitely generated subgroups. Solution Analogously, finitely generated subgroups are represented in the same kind of lattice structure with atoms and finite unions of generators. c) Can you obtain as a direct limit of finite abelian groups? Solution No, only torsion groups are obtainable. As finite abelian groups are all isomorphic to direct sums of /p k and we ll be taking a 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 some relations, the direct limit is always nonempty, unless the disjoint union itself consists of empty sets. Therefore all sets in the direct system must be 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 the direct sum of all groups and taking the quotient with the subgroup generated by all i α (g α ) i β (res α,β (g α )) we need to obtain all elements of the direct sum. Hence i β (res α,β (g α )) must be zero for all (α, β) Λ 1. So res α,β must kill all elements, we therefore have rather boring restriction morphisms. 2

3 5. What can you say about the direct limit of a direct system all of whose maps are injective? Surjective? Solution All the maps to the direct limit are injective (or surjective). In the injective case the direct limit contains information on all objects of the direct system, in the surjective case the direct limit contains the common information of all objects. 6. For n 0, let C n (x) denote a cyclic group of order n with generator x. Let p be a prime number. Let G be the direct limit of the following direct system of abelian groups: {0} = C p 0(x 0 ) C p 1(x 1 )... C p n(x n ) C p n+1(x n+1 )... (1) (where C p n(x n ) C p n+1(x n+1 ) takes x n px n+1 ). Preferably without resorting 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 big structure as every map is injective. As every step in the chain adds a finite number of elements and there are countable steps, we obtain a countable infinite group. As the construction of the direct limit consists of summing all objects of the direct system and taking a quotient, we get that every element of the 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 such that the term for C p n is nonzero but for C p n+1 is zero. The order of the finitely-generated subgroup is now limited by the order p n. Deduce that G has no proper infinite subgroup, and no maximal proper subgroup. Can either of these situations arise for subspaces of a vector space (using dimension instead of order)? Identify a realisation of G inside the unit circle S 1 (under multiplication). Solution The construction of the direct sum learns us that there must be an infinite number of nonzero summands, but as every C p n is trivially embedded in C p m for all m > n and every such instance is identified we cannot get to a proper infinite subgroup. Analogously there is no maximal proper subgroup. The only interesting case occurs in infinite-dimensional vector spaces, the finite-dimensional case is trivial (take the quotient of the space with the field). The obtained structure is the Prüfer group. 3

4 7. Consider the following direct system of abelian groups: fix r ; for all n let U n = and for n m let ρ m,n : U m U n be multiplication by r n m. Identify the lim as a subring of. Solution This is the localization at the ideal generated by r. In the case of r prime this is (p), in the other case we obtain all fractions with prime factors of r as denominators. 8. Interpret and prove: the direct limit of a system of exact sequences is exact. Solution For 0 A n B n C n 0 an exact sequence all the maps described in the definition of a direct limit are commuting by the properties of injective and surjective maps. 9. The notions of target and direct limit can be formulated without the restriction (a) of Definitions 3.1 and What difference does this make to the constructions? Find a system of abelian groups (in this generalised sense) with direct limit A B without having this abelian group appear in the system. Justify Remark Solution Just taking a quotient won t work, there are several components in the generalised direct system that must be handled appropriately. Take X :=]0, 1[ ]1, 2[ and the Euclidean topology with X removed from the topology. Now the direct system doesn t satisfy the conditions, but by taking a direct system on ]0, 1[ that produces A as its direct limit (in the former sense) and likewise on ]1, 2[ one that produces B we obtain a system of abelian groups in the generalised sense that produces A B. 10. Formulate the dual notions of inverse system and inverse limit lim (reverse the arrows). Solution As the exercise suggests, this is just a reversal of the arrows ρ α,β to U β U α, for α β. Find inverse systems: a) of finite sets whose lim is infinite; Solution Take (, ) and define A k = {0,..., k}. We get ρ i,j : A j A i the projection for i j. The direct limit is the set. b) of finite abelian groups whose lim is infinite; Solution The ring of p-adic integers p is given by lim /p n. c) of abelian groups whose lim is (without in the system). Solution Start from. Let p n denote the nth prime number. Now define G k to be the subgroup of in which all primes but p 1,..., p n are allowed as factors of the denumerator. The morphisms are the injections. Now is the inverse limit, it being the subgroup of in which all prime factors of denumerators are removed. 4

5 11. Verify that if (R α ) α Λ is a direct system of abelian groups such that each R α is a ring and all the ρ α,β are ring morphisms, then lim R α has a natural ring structure such that all the maps R β lim R α are ring morphisms. Solution The ring structure is preserved by the direct sum construction of the direct limit of abelian groups. As the subgroup H 1 is the construction is an ideal because the multiplicative structure is preserved by the injections, we obtain a quotient ring. 12. What are the stalks of the presheaf P 2 of 2.E? Solution We have (P 2 ) x0 = as every open subset U containing x 0 will have P 2 (U) =, with ρ α,β = id in the direct system. For (P 2 ) x with x x 0 we get {0}, the trivial group. We have x x 0 = ε > 0, so for the open set B(x, ε) [0, 1] the presheaf gives {0}, all values of the presheaf from there on are the same trivial group. 13. Construct a topological space X and a presheaf F of abelian groups on X with the properties: a) for any open U X : F(U) {0}; b) for all x X the stalk F x = {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 maps restricting everything to zero, but by the sheaf conditions this is impossible. 5

6 2 Exercises in Chapter 2 Exercise (1.9). Show that if G is an abelian sheaf, then G() = {0}, the trivial group. Solution By taking the empty cover of the empty set, the first part of the sequence that 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 embedded 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 P 1 from Example 1.2.E. Now define f : F G to be the trivial map for F(U) G(U), U X open and let f (X ): F(X ) G(X ) : n 2n. Similarly, g(x ): F(X ) G(X ) : n 3n. Now f and g agree on all stalks but as the monopresheaf fails (all restrictions are elements of the trivial group but the global behaviour differs) we have f 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 morphism of sections Γ (U, E) Γ (U, E) : σ id σ, which is locally an identity. This extends 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 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 and Γ f x : (Γ E) x Γ E x (5) f p 1 (x) : p 1 (x) p 1 (x) (6) are isomorphic maps. Solution This is nothing but Proposition

7 Exercise (3.9). Check the functorial properties L(f g) = L f Lg L(id) = id (7) Solution The map id x : F x F x is applied for all x and L(F) is the disjoint union of these stalks, so L(id) = id. The composition is handled as in Γ (f g). 7

8 3 Exercises at the end of Chapter 2 1. Let I = [0, 1]. Show that there is a unique (up to isomorphism) sheaf F on I with stalks: F 0 = F 1 =, F x = {0} if x I \ {0, 1}. Solution A presheaf has a unique sheafification, so we need to define a presheaf with these stalks and sheafify it. Do this by letting Γ (U, E) = if {0, 1} U and Γ (U, E) = {0} otherwise. Now the stalks agree and Γ (E) = F. What is Γ (I, F)? Solution It is, as it is composed of contributions of the stalks in 0 and 1. Let G be the constant sheaf on I. How many morphisms are there from F to G? From G to F? Solution To make the triangle in the definition of a (pre)sheaf morphism commute, observe that res U V is either the identity1 of the projection on {0} while res U V is the identity for all open sets. So for a morphism from F to G we have on option: the identity on U containing 0 or 1 and the canonical injection on the other open sets. For a morphism from G to F on the other hand we have total freedom for U not containing 0 or 1: whatever we do on these open sets doesn t matter as the 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 connected neighbourhoods); b) for any set A, the constant sheaf A X has Γ(U, A X ) = t U A for U open in 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, A X ) = locally constant functions U A (8) and as any covering of U falls apart in disjoint coverings of every connected component of U we obtain t U A. 2b 2c Trivial specialization. 1 Or some isomorphic map, the image of 1 defines the entire morphism and every morphism corresponds to a multiplication with a scalar. 8

9 2c 2a As Γ (U, {0, 1} X ) = t U {0, 1} we obtain a notion of indicator functions. They determine uniquely to which connected components an open subset of U belongs. When these conditions hold, what are the restriction maps in terms of the representation 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 components. 3. Let F be a presheaf on a space X, and let V be open in X. Then we can define 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 open covering of this set in V in the topology on X. Now the conditions for the sheaf 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 corresponding sheaf acts on V. Now Lemma 3.5(b) provides the necessary conditions, 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 F over A by Γ (A, F) = Γ (A, LF) (9) = sections of the continuous map p 1 (A) p A. (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 such that U A. (Colloquially, this says that a section of F over A extends uniquely into a small neighbourhood of A.) Solution A section over A is a section for p: p 1 (A) A, but continuity is defined in terms of open sets, so every section over A must be a section for some U A. By taking the direct limit we identify equal sections. 5. Let F be a sheaf on a space X and let (M i ) i I be a locally finite covering of X by closed sets (so that for each x X, i I x M i is finite). In the notation of Q4, suppose we are given a family (s i ) i I with i I : s i Γ (M i, F) and i, j I : s i = s j on M i M j. 9

10 Show that there is a unique s Γ (X, F) with i I : s = s i on M i. Solution By the previous exercise all sections on a closed M i are lifted uniquely to a small (open) neighbourhood and we ll take a finite intersection of opens. Now we can apply the glueing condition for sheaves and find a global 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 whose closed sets are the finite subsets of K. Define a sheaf of commutative unital rings on X as follows: for U open in X, U, let (U) = f L g, h K[t]: f = g h P U : h(p) 0. (11) If V U then (U) (V ) L and we take the inclusion as the restriction map res U V. Show that is a sheaf of rings on X. Solution As X is equipped with the cofinite topology, it is compact. Now every open covering reduces to a finite covering. Now the conditions for a sheaf are fulfilled: to every open set we ve assigned an algebra, restrictions are correct considering them as K-valued functions and the glueing conditions reduces to an argument on the poles of the section (when considered over X ). A better answer would be to state that the sections over U are defined as the intersection of the sections over the (finite) covering of U. It is the inverse limit of algebras realized as a pullback, which is exactly this intersection for our finite covering. Identify the stalk P of at P X as a subring of L and show that it is a local ring. What is its residue field and its field of fractions? Solution The stalk corresponds to P = f L g, h K[t]: f = g h h(p) 0. (12) Now the ideal of f such that f (P) = 0 is the unique maximal ideal because the functions for which f (P) 0 are invertible, hence not part of any maximal ideal. The quotient gives a residue field, which is an extension field of K. The corresponding field of fractions is just L. When does have non-polynomial global sections? That is, we certainly have K[t] Γ (X, ); when is the equality strict? Solution When K is not algebraically closed, there must be polynomials h K[t] such that h(p) 0 for all P X. For instance t in [t] provides a good denominator for non-polynomial global sections. 10

11 This is an equivalence: a non-polynomial global section must consist of a fraction of polynomials (exactly by the definition of the sections), so its denominator cannot have a zero. This only occurs when K is not algebraically closed (otherwise the denominator factors into linear components). We can consider f Γ (U, ) as a function on U, namely express f = g/h as in (11) and define for P U f (P) = g(p)/h(p) K. (13) Show that this defines a morphism φ : F where F is said to be the sheaf of K-valued functions on X (do this by giving K the indiscrete topology), and that putting (U) = Im φ(u) defines a sheaf with a morphism. Prove that is an isomorphism of sheaves. Hence we may regard 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 open set U. This is a sheaf by trivial reasons of gluing and restrictions: we ve got a pointwise definition of a function. For the isomorphism of sheaves: for we obtain an inverse map because functions are one-valued. For the isomorphism is obtained by the same argument. As g(p)/h(p) K this easily defines a good morphism. By removing enough points from an open U in order to determine enough values to uniquely define f on U we get a nice sheaf morphism. Now is a sheaf morphism because of this property. For K = show that is a subsheaf of the sheaf ω of analytic -valued functions on X =. Solution For K = we know that these K-valued functions correspond uniquely to rational functions, which are ω on a suitable open set (as they are meromorphic). 11

12 4 Exercises in Chapter 3 Exercise (1.3). Show that identities are unique. Solution Take B = A. If both id A and id A fulfill the conditions of an identity morphism, we have id A id A = id A by f id A = f (14) and therefore id A = id A. = id A by id A f = f (15) Exercise (1.8). Show that a natural transformation n is a natural equivalence if and only if A Ob C we have that n A is an isomorphism. Solution Just put the two diagrams from the definition next to eachother and chase the arrows: F(A) n A G(A) m A F(A) is also given by id F (A) and vice versa. The opposite 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 H g F f G = 0 then g factors uniquely as H g (16) Ker(f ) F Solution The factorization is uniquely determined by g, as it will map the sheaf G into 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 R f G g H = 0 then g factors uniquely as G g (17) PCok(f ) H Solution Analoguous to Exercise

13 Exercise (6.2). Formulate the universal property that you would like a concept of image to satisfy and verify that PIm and SIm do in the categories Presh/X and Shv/X. Solution Given a morphism f : F G, we want the image of f to be the object I such 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 monomorphism h : I G such that f = i f (i.e., it tries to act like our image) we have 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 respective cokernels we know that the images of f are mapped to zero, so the kernel of 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 sections over each open U is the image of f (U), while SIm(f ) is a sheaf whose stalk at each x X is the image of f x. 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 sequence which 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 (id F ) = id φ F because by definition of the direct image of a morphism φ idf (U) = idf φ 1 (U) = id φ F (U) (18) and id = id by the definition of the direct image (pre)sheaf. Now take F f G g H (pre)sheaf morphisms on continuous maps X φ1 Y φ2 Z. We now have φ 2 φ 1 g f (U) = g f φ 2 φ 1 1 (U) (19) = = by peeling of the definition of φ f. g f φ 1 1 φ 2 1 (U) (20) f φ 1 1 φ 2 g(u) (21) = φ 2 g φ1 f (U) (22) 13

14 5 Exercises at the end of Chapter 3 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 maps of sets f : A B such that f (a) = b. Show that P is a category with a zero object, kernels and cokernels, but in which not every epimorphism is a cokernel. 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 the base point and hence a unique map from our initial object is defined. The terminal object {x} in Sets remains unchanged as it allows for exactly one morphism (which preserves the base point). For kernels and cokernels we need a zero morphism, which in this case is the projection on the base point. For a map f : (A, a) (B, b) we have that 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 epimorphisms being surjective, while the cokernel removes the image except for the base point. 2. For X any topological space, show that the following presheaves of sets over X are in facts sheaves: a) Fixing an open V X, let singleton h V (U) = U V U V (23) for U open in X, with the unique restrictions. Solution Both the monopresheaf and the gluing condition are meaningless for U V, so they trivially hold. For U V both conditions are valid as there is only one possible section. Therefore all restrictions and gluings are well-defined. b) Let Ω(U) = W W X open and W U for U open in X with restriction: Ω(U) Ω(V ) : W W V. Interpreting presheaves on X as contravariant set-valued functors on, the category of open sets of X, show that the h V defined by 2a are the representable functors h V = Hom (, V ): U Hom (U, V ). (24) Solution First of all note that Ω(U) describes the (open) subspace topology induced by X on U. Now we have in any topology (which is just a lattice) that {i: U V } U V Hom (U, V ) = (25) U V 14

15 which corresponds to h V = Hom (, V ). Interpret the Yoneda lemma as saying that Hom Presh/X (h V, F) F(V ) (26) for any presheaf F on X. Putting F = h U, this shows that V h V is a full and faithful embedding of into Presh/X. Solution We have that is locally small (as each set of homomorphisms contains either zero or one morphism, which is quite small), so by the (contravariant) Yoneda lemma we can assign to each object V a functor h V = Hom(, V ) to Sets. Now we have that for each object V the natural transformations from the presheaf (of representable functor) h V to a presheaf F (which is just a functor, we are working in an functor category) correspond bijectively to F(V ). I.e., Hom Presh/X (h V, F) 1:1 F(V ) (27) 4. Let F, G be presheaves of sets on a topological space X. Define a new presheaf Hom(F, G) with Hom(F, G)(U) := Hom Sets (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 restriction res U V (f ) we get by commutativity in the diagram for sheaf morphisms a good section in Hom(F, G)(V ), compatible with all restrictions, for V U open. Hom(F, G) monopresheaf Given the situation for the monopresheaf condition, if s s, both being maps F(U) G(U) they must disagree on some x U, i.e., s(u) s(u ). But u U λ for some λ Λ, so res U U λ (s)(x) res U U λ (s )(u), a contradiction. Hom(F, G) glues Define the section over U by its values in the U λ, the obvious restrictions are satisfied and we get a section glued from its components. Prove that this construction has the following universal property: for F, G, H presheaves (respectively sheaves) of sets on X, there is a bijection Hom H, Hom(F, G) = Hom (H F, G) (29) 15

16 natural in F, G and H. Here H F is the product object provided by 3(d); it is constructed pointwise. Solution Take Hom(H, Hom(F, G)), that is for U X open we have Γ (U, ): H(U) Hom(F, G)(U) = Hom Sets (F(U), G(U)) h fh : F(U) G(U) : x f h (x). (30) This is now mapped to Hom(H F, G) by mapping each Γ (U, ) to the map (H F) (U) = H(U) F(U) G(U) (h, f ) (h)(f ) (31) surjectivity We can uniquely define an Hom(H, Hom(F, G)) by this construction, so the map is surjective. In more detail, define Γ (U, ) H(U) Hom(F, G)(U) (32) h f g and this provides a good preimage. injectivity Assume 1 and 2 in Hom(H, Hom(G, F)) are mapped to the same element in Hom(H F, G). Then by the construction above the inverse image of this element corresponds to a unique element in the domain, a contradiction. Show that, if we want the property (29) to hold, then the definition of Hom is forced on us, for presheaves at least. Solution Taking H = h U, now we want Hom(h U, Hom(F, G)) = Hom(h U F, G) (33) to hold. But on all V U this reduces to Hom Sets (F(U), G(U)) because in 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 subobject classifier, that is, an object Ω with the equivalent properties (prove their equivalence): 16

17 a) there is a natural bijection Hom(F, Ω) = {subobjects of F} (35) for any object F; b) Ω has a special subobject 1 t Ω such that any morphism G F is the pullback of t over a unique morphism F Ω (called the classifying map of G F). Proof. As 1 Ω is a monomorphism, we have that the realization of the pullback over F Ω is again a monomorphism. As pullbacks are unique (up to isomorphism) we find the bijection (and vice versa every map g : F Ω provides 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 = h U, by the Yoneda lemma we have Hom(h U, Ω) = Ω(U) 1:1 subobjects of h U. (36) But a subobject of h U is an h V for V U open, so by the full and faithful embedding of into Presh/X, we have Ω(U) = V X V open, V U. (37) For a general presheaf F we now find Ω(U) = { (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 V X open ρ U V (x) G(V ). (39) This defines a bijection between morphisms from F to Ω and subobjects of F as the map exactly describes how a subobject is constructed, giving for 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 is a 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 characteristic function 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 the bijection found concerns the characteristic functions from F to {0, 1}, hence they define the subsets. The pullback is just the monomorphism. 17

18 6. Since any category of presheaves is a functor category, we can define it for categories other than those of open sets of a topological space. For op any category, let Presh/ be the category Sets of contravariant functors Sets (and natural transformations). As a special case, any group G can be regarded as a category with one object whose endomorphisms are the elements of G, with composition defined as in G (hence every morphism is an isomorphism). Show that the category Presh/G can be regarded as the category of sets-with-a-g op -action, that is the category of permutation representations of G. Solution As F Presh/G is a contravariant functor G Sets and by definition 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, not just a monoid) we have F(End( G)) = Aut F( G), i.e., all bijections or permutations on F( G). Now every g G defines an (invertible) action on the 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 abelian groups 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 a monopresheaf using Corollary 3.7. For the glueing condition: we can glue a section s in Q(U). Now the (s λ ) λ are mapped to 0 in R(U λ ) and by the monopresheaf condition with 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 monopresheaf. By the exactness we find in Q(U λ ) that the cosets of ρ U U λ (s) and 0 are equal by lifting the quotient. Hence ρ U U λ (s)+p λ = p λ, both p λ and p λ in the embedding of P(U λ) in Q(U λ ). Now we can glue the p λ and p together, from which s = 0 follows by λ the monopresheaf condition on Q. 10. A sheaf F of abelian groups on X is called locally free if and only if each point x X has an open neighbourhood U in X such that the sheaf F U is 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 a well-defined rank. Solution As it is locally free, the rank of the stalk is extended in a small neighbourhood. But as X is connected this procedure extends to all of X. 18

19 Show by example that a locally free sheaf on X need not be isomorphic to a constant sheaf even if X is connected, and that φ does not preserve the property of being locally free. Solution Take X = {0, 1, 2} with topology = {{0, 1}, {1}, {1, 2}, X } and define F to be the sheaf with sections Γ ({0, 1}, F) = (40) Γ ({1, 2}, F) = (41) Γ ({1}, F) = (42) Γ ({0, 1, 2}, F) = (43) with restriction morphisms the projection on the first term. Now take for x X any open U X, then F U = U, but F is not isomorphic to any constant sheaf. Prove however that the inverse image of a locally free sheaf of rank n is locally free and of the same rank. Solution The inverse image f is given locally by the direct limit of open sets containing the inverse image of the continuous map f, but this direct limit consists of data of a locally free sheaf, hence it is locally free. 6 Exercises at the end of Chapter 4 3. Draw a picture of Spec [t]. Solution In Figure 1 my own L A TEX version of Mumford s Spec [t] is given. 4. Show that for any ring R, Spec R is compact. Solution We start from Spec R = i I U i. We have for all i I an open covering of U i by distinguised opens, i.e., sets of the form D(f ), as the topology is given by a basis, closed under intersection. Hence, we obtained an open covering of Spec R using more sets such that every U i can is exactly covered. Now define S the set of all these representatives of distinguised opens. The set S cannot be contained in any proper ideal or we would find a maximal ideal not yet covered, so S generates all of R. Now write unity as a finite linear combination. The corresponding distinguised opens cover Spec R, so we have reduced the covering to a finite covering. 9. Algebraic curves 19

20 Topological version Show that the subset of 2 given by C = (x, y) y 2 = (x + 1)x(x 1) (44) is a manifold (topological, differentiable or analytic as you wish), whereas that given by the equation y 2 = x 2 (x + 1) is not. Solution In Figure 2 an impression in 2 is given. In the first case one component is homeomorphic (or diffeomorphic as you wish) to a circle while the other is homeomorphic to. In the second case though, the self-intersection in the origin creates a problem: there is no correct definition of the tangent. Or if you are looking for a topological argument: removing the origin creates three components, which is impossible for a manifold, given its local Euclidean structure. Algebraic version Let K be any field and f (x, y) K[x, y] = R. By Q6 the morphism R R/(f ) identifies the space Spec R/(f ) with a closed subspace C of Spec R = 2 K = X say. Find the sheaf of ideals I in X C such that, letting be the quotient of X C by I, the morphism Spec R/(f ) (C, ) (45) is an isomorphism of ringed spaces (and so of affine schemes). Putting f (x, y) = y 2 x 3 + x or y 2 x 3 x 2 we get the algebraic analogues of the curves of part (a). What is different about the two cases? Solution The local ring in the origin is different: it shows the occurence of two tangent lines. It is a singular point. Other curves for your amusement: Solution y 2 = x 2 (x 1) (46) y 2 = x 3 (47) In Figure 3 the embedding in 2 is given. We see just one component containing nothing but regular points while the curve y 2 x 3 has a cusp: there is a multiple tangent in the origin. 20

21 [(2)] [(3)] [(5)] [(0)] generic point [(2, X + 1)] [(3, X + 2)] [(5, X + 4)] [(5, X + 3)] (X 2 + 1) [(5, X + 2)] [(3, X + 1)] [(5, X + 1)] [(2, X )] [(3, X )] [(5, X )] (X ) ((2)) ((3)) ((5)) ((7))... Figure 1: Mumford s treasure map y y x x (a) y 2 = (x + 1)x(x 1) (b) y 2 = x 2 (x + 1) Figure 2: Plots 21

22 y y x x (a) y 2 = x 2 (x 1) (b) y 2 = x 3 Figure 3: Plots 22

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