# Sheaves. S. Encinas. January 22, 2005 U V. F(U) F(V ) s s V. = s j Ui Uj there exists a unique section s F(U) such that s Ui = s i.

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1 Sheaves. S. Encinas January 22, 2005 Definition 1. Let X be a topological space. A presheaf over X is a functor F : Op(X) op Sets, such that F( ) = { }. Where Sets is the category of sets, { } denotes a set with one element and Op(X) is the category with objects the open sets of X and arrows the inclusions. Op(X) op denotes the oposite category: same objects and reversed arrows. A morphism of presheaves is a morphism in the functor category. F : Op(X) op Sets U F(U) = Γ(U, F) Every element of s F(U) is called a section of F in U. Γ(F, U) = F(U). If V U are two open sets, the restriction morphism is: U V F(U) F(V ) s s V In the literature is used also the notation Definition 2. A presheaf F is a sheaf if the following condition holds for every open set U: Given any open covering {U i } i I of U and given sections s i F(U i ) such that s i Ui U j = s j Ui Uj there exists a unique section s F(U) such that s Ui = s i. then One may define in a similar way presheave and sheaves of Groups, Rings, Modules, just by changing the category of sets by the corresponding category. Exercise 3. Let X be an algebriac variety over a field k. We define a sheaf O on X: For every open U, O(U) is the set of regular functions defined in U k. Recall that regular functions are the functions which are locally a quotient of polynomials (of the same degree if X P n ). It is obvious that O is a presheaf if we consider the usual restrictions of functions. Prove that O is a sheaf. Hint: By patching one obtains a function, which is a regular function by definition. Example 4. A presheaf which is not a sheaf. Let X be a topological space and let A be a fixed set. Define F(U) = A for every open set U. Is is trivial that F is a presheaf, the constant presheaf. But F is not a sheaf. Let U = U 1 U 2 be an open set with two connected components, U 1 U 2 =, Consider two sections s i A = F(U i ), i = 1, 2. If s 1 s 2 there is not any section s A = F(U) with s Ui = s i. 1

2 Definition 5. We recall the definitions of limit and colimit in a category C. Given objects {C i } i I and morphisms {f ij : C i C j } (i,j) Λ, with Λ I I. One may define the limit and the colomit of the C i s (w.r.t the morphisms f ij ): The limit lim i I C i, if it exists, is an object C together with morphisms f i : C C i such that: For every object A and morphisms g i : A C i such that f ij g i = g j for every (i, j) Λ, there exists a unique morphism g : A C with g i = f i g, for any i I. A C i g i g f i C f ij f j g j C j The colimit lim i I C i, if it exsits, is an object C together with morphisms f i : C i C such that: For every object A and morphisms g i : C i A such that g j f ij = g i for every (i, j) Λ, there exists a unique morphism g : A C with g i = gf i, for any i I. C i g i f i f ij C g f j g j C j A We say that the limit is filtered if for every (i, j) Λ there is a k I with (k, i), (k, j) Λ. The colimit is cofiltered if for every (i, j) Λ there is a k I with (i, k), (j, k) Λ. A special case of limit and colimit is when Λ =, we do not have morphisms f ij. In this case the limit is called product and the colimit is the coproduct. Definition 6. Let F be a presheaf and let x X. The stalk of F at the point x is F x = lim F(U) U x The stalk F x contains all the germs of sections at the point x. Every element s F x is an equivalence class, a representant is a pair (s, U) where s F(U) and x U. Two representants (s 1, U 1 ), (s 2, U 2 ) are equivalent if s 1 U1 U 2 = s 2 U1 U 2 in F(U 1 U 2 ). If s F(U) is a section and x U, the natural map F(U) F x sends s to the germ, which we will denote by s x. Exercise 7. Let F and G be two sheaves on X. If ϕ : F G is morphism of sheaves then prove that for any x X it induces naturally a morphism on the stalks ϕ x : F x G x. Proposition 8. Let F and G be two sheaves on X. A morphism of sheaves ϕ : F G is an isomorphism if and only if for every x X the induced morphism ϕ x : F x G x is an isomorphism. 2

3 Proof. If ϕ is isomorphism, there is an inverse ϕ 1 and then for every x X ϕ 1 x is the inverse of ϕ x. Conversely, assume now that ϕ x is an isomorphism for all x X. Let U X be an open set. First we prove that ϕ(u) is injective. Consider two sections s 1, s 2 F(U) such that t = ϕ(u)(s 1 ) = ϕ(u)(s 2 ). For any point x we have ϕ x (s 1,x ) = ϕ x (s 2,x ) = t x so that by hypothesis s 1,x = s 2,x for all x U. For every x U there is an open set x V x U such that s 1 Vx = s 2 Vx. Now {V x } x U is a covering of U and by the sheaf property s 1 = s 2. Now we construct the inverse ϕ 1 (U) : G(U) F(U) for any open set U. Let t G(U) and consider the germ t x G x at every x U. Let s x = ϕ 1 x (t x ), the germ s x has a representant s(x) F(V x ) for some open x V x U. The sections ϕ(v x )(s(x)) and t Vx represent the same germ t x at x, so that we may assume that V x is small enough such that ϕ(v x )(s(x)) = t Vx. We have section s(x) F(V x ), where {V x } x U is a covering of U. We want to prove that they patch to a section s F(U). If x, y U the restrictions s(x) Vx Vy and s(y) Vx Vy are such that ) ) ϕ(v x V y ) (s(x) Vx Vy = ϕ(v x V y ) (s(y) Vx Vy We have already prove that ϕ(v x V y ) is injective, so that s(x) Vx V y = s(y) Vx V y and the sections {s(x)} x U patch to a unique section s F(U). We have defined maps G(U) F(U) for U Op(X). Exercise: check that those maps are (ϕ(u)) 1 and they define a map of sheaves G F inverse of ϕ. Definition 9. Let F be a presheaf. The associated sheaf to F is a sheaf F + together with a morphism of presheaves θ : F F + such that: For any sheaf G and any morphism of presheaves ϕ : F G there is a unique morphism ψ : F + G with ϕ = ψθ. ϕ F G θ ψ Proposition 10. The sheaf associated to a presheaf exists and it is unique up to isomorphism. Proof. Let E = x X F x. Define F + (U) as the set of maps s : U E such that s(x) F x, for all x U. F + s is locally given by sections of F. More precisely for every x U there is a neighborhood x V U and a section t F(V ) such that s(y) = t y for all y V. Exercise 11. Check that F + is a sheaf and it satisfies the universal property of the definition 9. Exercise 12. Let F be a sheaf. Prove that the associated sheaf to F is F + = F. Exercise 13. Consider the presheaf of 4. Prove that the associated sheaf F + is the following: If U is an open set with r connected components, then Definition 14. Let f : X Y a continuous map. F + (U) = A r A 3

4 If F is a sheaf on X, we define the direct image sheaf f F: f F(V ) = F ( f 1 (V ) ), V Op(Y ) If G is a sheaf on Y, we define the sheaf inverse image f 1 (G) as the associated sheaf to the presheaf: U lim G(V ), U Op(X) Example 15. A sheaf G such that the inverse image presheaf is not a sheaf. Assume that U = U 1 U 2 is a union of two connected components and W = f(u 1 ) = f(u 2 ) is an open set of Y. f 1 G(U i ) = lim G(V ) = G(f(U i )) = G(W ), i = 1, 2 V f(u i ) lim G(V ) = G(W ) f 1 G(U) = G(W ) G(W ) Exercise 16. Construct explicitly the previous example. Proposition 17. Let f : X Y be a continuous map. 1. If F is a sheaf on X, there is a natural morphism of sheaves on X, f 1 f F F. 2. If G is a sheaf on Y, there is a natural morphism of sheaves on Y, G f f 1 G. 3. Let F be a sheaf on X and G be a sheaf on Y. There is a natural bijection of sets: Hom X (f 1 G, F) = Hom Y (G, f F) Proof. 1. If U Op(X), f 1 f F(U) = lim f F(V ) = lim F(f 1 (V )) F(U) where the last arrow comes from f 1 (V ) U. Then we have a map of presheaves and the universal property of 9 gives the result. 2. Let V Op(Y ), G(V ) lim G(W ) = f 1 G(f 1 (V )) = f f 1 G(V ) W f(f 1 (V )) the first arrow comes from V f(f 1 (V )). This map of presheaves induces a map of sheaves. 3. It follows from the previous results. Definition 18. An étalé space over a topological space X is a topological space E togheter with a morphism Π : E X such that for any point ξ E there exists a neighborhood W E such that Π W defines an homeomorphism to an open set of X. A section of the étalé space is a continuous map s : U E defined in an open set U of X such that Π s = Id U. 4

5 It is easy to check that the sections of the étalé space form a sheaf on X, which we will denoted by F(E). Proposition 19. Let F be a presheaf. Define E = x X F x and Π : E X the natural projection. For every open set U, and every section s F(U), we define s : U E as s(x) = s x. Give to E the strongest topology such that the maps s are continuous. Then E is an étalé space. Moreover the sheaf F(E) is the sheaf F + associated to the presheaf F. Proof. Exercise. Exercise 20. The category of sheaves on X is equivalent to the category of étalé spaces over X. 5

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