# SOME OPERATIONS ON SHEAVES

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1 SOME OPERATIONS ON SHEAVES R. VIRK Contents 1. Pushforward 1 2. Pullback 3 3. The adjunction (f 1, f ) 4 4. Support of a sheaf 5 5. Extension by zero 5 6. The adjunction (j!, j ) 6 7. Sections with support in a closed subspace 7 8. The adjunction (i, i! ) 7 9. Recollement Mayer-Vietoris sequences Gluing sheaves Additional exercises The adjunction (f 1, f ) revisited 11 All sheaves are assumed to be abelian sheaves. For a space X we write Sh(X) for the category of sheaves on X. At first reading I recommend that you skip straight to 9 and 10. You may also want to glance at the section on adjoint functors in the notes for week 3. Further, when you do start reading the sections in this set of notes sequentially, keep a copy of 12 on the side and try to do these additional exercises as you go along. 1. Pushforward 1.1. Let f : X Y be a map between spaces and let F be a sheaf on X. Define a presheaf f F on Y by f F (V ) = F (f 1 (V )) for an open subset V of Y. If U is an open subset of V, define the restriction map f F (V ) f F (U) by s res f 1 (V ) f 1 (U) (s). The presheaf f F is a sheaf If α: F G is a morphism of sheaves on X, define a morphism f (α): f F f G as follows: a section of f F over an open subset V of X is, by definition, a section s F (f 1 (V )) and f (α) is given by mapping s to α f 1 (V )(s) which is a section of G(f 1 (V ). The latter is, by definition a section of f G(V ). Consequently, we obtain a functor f : Sh(X) Sh(Y ). The functor f is called the pushforward or the direct image along the map f. 1

2 2 R. VIRK 1.3. It is clear that if g : Y Z is another map between spaces, then g f = (gf) Example. Let X be a space and let a: X pt be the obvious map. If F is a sheaf on X, then a F = F (X) = Γ(X; F ). In this special case a is also called the global sections functor. The reason for this terminology should be obvious Example. Let X be a space and let x X. Then the pushforward of the constant sheaf, Z x on {x}, along the inclusion map i: {x} X, is given by { Z if x U; i Z x (U) = 0 otherwise, where U is an open subset of X. The sheaf i Z x is called the skyscraper sheaf supported on x Example. More generally, if f : X Y is the inclusion of a subspace and F is a sheaf on X, then f F (U) = F (U X), for open subsets U of Y Proposition. Let f : X Y be a map between spaces. Let 0 F G H be an exact sequence of sheaves on X. Then the sequence is exact. 0 f F f G f H Proof. As 0 F G H is exact, the sequence 0 F (U) G(U) H(U) is exact for every open subset U of X. The result follows Exercise. Show, via an example, that if 0 F G H 0 is an exact sequence of sheaves on a space X, then it is not necessarily true that 0 f F f G f H 0 is exact Proposition. Let i: Z X be the inclusion of a closed subspace. Then the functor i is exact. That is, if 0 F G H 0 is an exact sequence of sheaves on Z, then is exact. 0 i F i G i H 0 Proof. Courtesy of the previous result, we only need to show that i G i H 0 is exact. Let z Z and let s be a section of i H over an open neighborhood U of z. Then s is a section of H over U Z. As G H 0 is exact, there is some t z G z that maps to s z. It follows that (i G) z (i H) z 0 is exact. If x X Z, then the stalks (i G) x and (i H) x vanish. Hence, i G i H 0 is exact at all stalks Exercise. Is the analogue of the above result true for inclusions of open subspaces? The above discussion on exactness is summarized by saying that the functor f is left exact but not generally exact. However, if f = i is the inclusion of a closed subspace, then i is exact.

3 SOME OPERATIONS ON SHEAVES 3 2. Pullback 2.1. Let f : X Y be a map between spaces and let F be a sheaf on Y. Let U be an open subset of X. Define a presheaf f t F on X via the assignment Γ(U; f t F ) = lim F (V ), V f(u) where V runs through the open neighborhoods of f(u) and the maps in the limit diagram are the restriction maps (I am going to stop repeating this about the maps, it should be clear by now that the maps in limit diagrams of this form are going to be the restriction maps. Constantly repeating this is starting to make for some very tedious exposition). The restriction maps for this presheaf are defined in the obvious way. Note: if f(u) is open in Y, then lim V f(u) F (V ) = F (f(u)). The presheaf f t F is usually not a sheaf. The sheafification of this presheaf, denoted f 1 F, defines a functor f 1 : Sh(Y) Sh(X), called the pullback or the inverse image along the map f. The functor f 1 is defined on morphisms in the evident way Exercise. Find an example demonstrating that the presheaf f t F is not necessarily a sheaf Example. Let j : U X be the inclusion of an open subset and let F be a sheaf on X. Then the assignment V lim F (V ) = F (V ), V V V open in U (and hence open in X), is already a sheaf. Thus, j 1 F (V ) = F (V ) The definition of f 1 implies that for x X, 2.5. Exercise. Verify this assertion. (f 1 F ) x = F f(x). (2.4.1) 2.6. Let g : Y Z be another map of spaces. Then one checks, either directly from the definitions or using the results of the next section, that (gf) = f g Exercise. Verify this Remark. Strictly speaking, the equality above is really a canonical isomorphism. However, this canonical isomorphism is coherent : there is no ambiguity in pulling back along three composable maps, all the diagrams obtained from these canonical isomorphisms commute, etc. So nothing is lost by replacing the isomorphism with equality Example. Let F be a sheaf on a space X, let x X and let i: {x} X be the inclusion map. Then i 1 F = F x Example. Let F be a sheaf on a space X and let f : W X be the inclusion of a subspace. Then the sheaf f 1 F is called the restriction of F to W. This sheaf is often denoted F W Example. Let a: X pt be the obvious map. Then a 1 Z pt is canonically isomorphic to Z X.

4 4 R. VIRK Proposition. Let f : X Y be a map between spaces and let 0 F G H 0 be an exact sequence of sequence of sheaves on Y. Then the sequence is exact. 0 f 1 F f 1 G f 1 H 0 Proof. It suffices to check exactness at stalks. This follows from (2.4.1) In other words, f 1 is an exact functor Warning. In the literature, the functor f 1 is sometimes denoted by f. Some care needs to be exercised to avoid confusion, since the functor f is a different beast from pullback maps on cohomology groups. To further add to the confusion, if one is dealing with sheaves in algebraic geometry (à la coherent/quasi-coherent sheaves) then f is what we are calling f 1 composed with tensoring with the structure sheaf (whatever that may be). On the same note, f should not be confused with pushforward maps on homology groups. 3. The adjunction (f 1, f ) 3.1. Let f : X Y be a map between spaces. Let F be a sheaf on Y and let U be an open subset of Y. Then Γ(U; f f 1 F ) = Γ(f 1 (U); f 1 F ). Clearly U f(f 1 (U)). So, using the defining property of a colimit we obtain a map F (U) lim V f(f 1 (U)) F (V ) = Γ(f 1 (U), f t F ). Composing this with the canonical morphism from a presheaf to its sheafification, we obtain a morphism of sheaves η F : F f f 1 F Note that a section s F (U) maps to 0 under η F if and only if there is an open subset V of U containing f(f 1 (U)) such that s V = The family of morphisms η F, F Sh(Y ), defines a natural transformation η : id f f Example. Let j : U X be the inclusion of an open subset. Then we have the following description of the unit map η : id j j 1. Let F be a sheaf on X and let V be an open subset of X. Then j j 1 F (V ) = F (U V ) and if s F (V ) then η F maps s to s U V Let G be a sheaf on X and let V be an open subset of X. Then f t f G(V ) = lim f G(V ) = lim G(f 1 (V )), V f(v ) V f(v ) where V runs through open subsets of Y containing f(v ). Certainly, if V contains f(v ), then f 1 (V ) contains V. Hence, using the universal property of the colimit we obtain a canonical map f t f G(V ) G(V ). This gives a morphism from the presheaf f t f F to G. Using the universal property of sheafification we obtain a morphism of sheaves ε G : f 1 f G G.

5 SOME OPERATIONS ON SHEAVES The morphism ε G is easy to describe at the level of stalks. Let x X. Then (f 1 f G) x = (f G) f(x) = lim G(f 1 (V )). V f(x) The map (f 1 f G) x G x induced by ε G is the evident one given by the universal property of the colimit In particular, if f : X Y is the inclusion of a subspace, and F is a sheaf on X, then ε F : f 1 f F F is an isomorphism The family of morphisms ε G, G Sh(X), defines a natural transformation ε: f 1 f id. With a bit of patience one checks that the compositions f G η f G f f 1 f G f (ε G) f G and f 1 F f 1 (η F ) f 1 f f 1 F ε f 1 F f 1 F are equal to the identity on f G and f 1 F respectively. In other words, f 1 is left adjoint to f Exercise. Check this! 4. Support of a sheaf 4.1. Let F be a sheaf on a space X. Let U X be an open subset. Recall that the support of a section s F (U) is the complement in U of the union of open sets on which s restricts to 0. The support of F, denoted supp F, is defined to be the complement in X of the union of all open subsets U of X such that F U = It is clear that if x supp F then F x = 0. However, the converse is not true Exercise. Find an example demonstrating that the reverse implication is false Proposition. We have supp F = {x X F x 0}. Proof. Let x supp F. Then for every open neighborhood U of x the restriction F U is non-zero. In particular, every open neighborhood of x contains a point at which the stalk of F is non-zero. Hence, x {x X F x 0}. Conversely, let y {x X F x 0}. Then every open neighborhood of y contains a point at which the stalk of F does not vanish. Hence, F U 0 for every open neigborhood U of x. Thus, y supp F Exercise. Let j : U X be the inclusion of an open subset. Let F be a sheaf on U. Is it true that supp j F is contained in U? What about if we replace the open inclusion by a closed inclusion? 4.6. Exercise. Let Y be a space and let X Y be a subspace. Let Sh X (Y ) be the category of sheaves on Y with support contained in X. Prove that the categories Sh(X) and Sh X (Y ) are equivalent. 5. Extension by zero 5.1. Let i: Y X be the inclusion of a closed subspace and F a sheaf on Y. Then the sheaf i F on X is the unique (up to isomorphism) sheaf F Y on X such that F Y Y = F and F Y X Y = 0. In particular, (F Y ) x = 0 if x Y, and (F Y ) y = F y if y Y The above statements are not true if we replace closed by open. We will now rectify this situation.

6 6 R. VIRK 5.3. Let j : U X be the inclusion of an open subset and let F be a sheaf on U. Define a sheaf j! F on X by { F (V ) if V U; j! F (V ) = 0 otherwise, where V is an open subset of X. Restriction maps are defined in the obvious way. Then we have a functor j! : Sh(U) Sh(X), with j! defined on morphisms in the evident way. The sheaf j! F is called the extension by zero of F to X. It is clear that if j : U U is the inclusion of an open subset, then (j j)! = j! j! The sheaf j! F is the unique (up to isomorphism) sheaf on F U on X such that F U U = F and F U X U = 0. The sheaf F U = j! F is called the extension by zero of F to X Proposition. The functor j! is exact. That is, if 0 F G H 0 is an exact sequence of sheaves on U, then is an exact sequence of sheaves on X. 0 j! F j! G j! H 0 Proof. Exactness at the level of stalks is clear. 6. The adjunction (j!, j ) 6.1. Let X be a space, let j : U X be the inclusion of an open subspace and let F be a sheaf on U, and let V U be an open subset. Then j t j! F (V ) = F (V ). As j is an open inclusion, the presheaf j t j! F is a sheaf. Hence, we obtain a canonical isomorphism η F : F j j! Now let G be a sheaf on X, and let V be an open subset of X. Then { j! j t G(V G(V ) if V U; ) = 0 otherwise. Consequently, we obtain a map j! j t F F. Now j t F is canonically isomorphic to j F. So we obtain a canonical morphism ε F : j! j F F. The families η F, F Sh(U), and ε G, G Sh(X), define natural transformations 6.3. The compositions η : id j j! and ε: j! j id. j G η j G j j! j G j (ε G ) j G and j! F j!(η F ) j! j j! F ε j! F j! F are equal to the identity on j G and j! F respectively. In other words, j! is left adjoint to j Exercise. Verify this assertion.

7 SOME OPERATIONS ON SHEAVES 7 7. Sections with support in a closed subspace 7.1. Let i: Z X be the inclusion of a closed subspace and let F be a sheaf on X. Define a sheaf F by F (U) = {s F (U) supp s Z}. (7.1.1) As usual, restriction maps are defined in the obvious way Exercise. Show that F is indeed a sheaf Define i! : Sh(X) Sh(Z), F i F Proposition. The functor i! is left exact. That is, if 0 F G H is an exact sequence of sheaves on X, then is exact Exercise. Verify this. 0 i! F i! G i! H 8. The adjunction (i, i! ) 8.1. Let i: Z X be the inclusion of a closed subspace. Let F be a sheaf on X and let F be as in (7.1.1). Let U X be an open subset. Then Γ(U; i i! F ) = Γ(U Z; i! F ) = Γ(U Z; i F ). Now Γ(U Z; i t F ) = lim V U Z F (V ) = F (U Z). Hence, we obtain a canonical isomorphism F i i! F. Furthermore, there is an obvious (injective) map F F. Consequently, we obtain a canonical morphism ε F : i i! F F Now let G be a sheaf on Z and let (i G) be as in (7.1.1). Then (i G) = i G. Consequently, if V is an open subset of Z, then Γ(V ; i t (i G) ) = lim U V Γ(U; i G) = lim G(U Z) = G(V ). U V Hence, we obtain a map G i t (i G) which induces a canonical isomorphism η G : G i! i G. The families η G, G Sh(Z), and ε F, F Sh(X), define natural transformations 8.3. The compositions η : id i! i and ε: i i! id. i! F η i! F i! i i! F i! (ε F ) i! F and i G i (η G) i i! i G ε i G i G are equal to the identity on i! F and i G respectively. In other words, i is left adjoint to i! Exercise. Verify this assertion.

8 8 R. VIRK 9. Recollement 9.1. Exercise. Meditate on why the title of this section is what it is The myriad relations between pushing forward and pulling back can be a bit overwhelming at first. Here s a quick summary to help you keep track of some of the important ones For every map f : X Y between spaces there are functors f : Sh(X) Sh(Y ) pushforward (left exact); f 1 : Sh(Y ) Sh(X) pullback (exact) Let j : U X be the inclusion of an open subspace and let i: Z X be the inclusion of the closed complement Z = X U. Then the categories Sh(U), Sh(X) and Sh(Z) are related by the following functors: j : Sh(U) Sh(X) pushforward (left exact); j! : Sh(U) Sh(X) extension by 0 (exact); j : Sh(X) Sh(U) restriction (exact); i : Sh(Z) Sh(X) pushforward (exact); i : Sh(X) Sh(Z) restriction (exact); i! : Sh(X) Sh(Z) sections with support in Z (left exact) We have adjunctions: (j!, j, j ) and (i, i, i! ), where (x, y, z) means x is left adjoint to y and y is left adjoint to z Further, we have the identities: j i = 0, i j! = 0 and i! j = The adjunction maps give canonical isomorphisms: i i id i! i and j j id j j! The adjunction maps also give exact sequences 0 j! j F F i i F 0 and 0 i i! F F j j F, for every F Sh(X). 10. Mayer-Vietoris sequences Exercise. Meditate on why the title of this section is what it is Let j : U X be the incluson of an open subspace and let i: Z X be the inclusion of a closed subspace. Note: we are not assuming Z = X U. Let F Sh(X). Define Γ U (F ) = j j F, Γ Z (F ) = i i! F, F U = j! j F, F Z = i i F.

9 SOME OPERATIONS ON SHEAVES Now let Z 1, Z 2 X be closed subsets that cover X. Then one obtains a short exact sequence 0 F + F Z1 F Z2 FZ1 Z 2 0, where + = ( η1 η 2 ), = ( η 1 η 2 ), and η i : F F Zi, η i : F Z i F Z1 Z 2, i {1, 2}, are the adjunction maps Similarly, if U 1, U 2 X are open subsets that cover X, then we have a short exact sequence where + = ( ε 1 ε 2 ), = are the adjunction maps. 0 F U1 U 2 + FU1 F U2 F 0, ( ε 1 ε Analogously, we have exact sequences and ), and ε i : F U1 U 2 F Ui, ε i : F U i F, i {1, 2}, 0 F + Γ U1 (F ) Γ U2 (F ) Γ U1 U 2 (F ) 0 Γ Z1 Z 2 (F ) + Γ Z1 (F ) Γ Z2 (F ) F, where the maps + and are induced by adjunction maps Exercise. Verify that these sequences are indeed exact. Hint: This can be done extremely quickly if you look at stalks. 11. Gluing sheaves Let X be a space and let i I U i = X be an open cover of X. Set U ij = U i U j and set U ijk = U i U j U k for i, j, k I Consider a sheaf F on X. Set F i = F Ui, for i, j I. Then we have canonical isomorphisms θ i : F Uij Fi Uij, for all i, j I. Let θ ji = θ j θ 1. Then (i) θ ii = id Fi, for all i I; (ii) (θ ij Uijk ) (θ jk Uijk ) = θ ik Uijk, for all i, j, k I. In fact, one can reconstruct F from the above data: Proposition. Let i I U i = X be an open cover of X. Assume to be given the following data: (i) a sheaf F i on U i for each i I; (ii) for each pair (i, j) I I an isomorphism θ ji : F i Uij Fj Uij. These isomorphisms satisfying: θ ii = id Fi and (θ ij ) Uijk (θ jk ) Uijk = θ ik Uijk for all i, j, k I. Then there exists a sheaf F on X, and isomorphisms f i : F Ui Fi, such that θ ij f j Uij = f i Uij. Moreover, the family (F, {f i } i I ) is unique up to canonical isomorphism. Proof sketch. For each open set U of X, define F (U) to be the submodule of i I F i(u U i ) consisting of families (s i ) i I such that for any (i, j) I I, θ ji (s i U Uji ) = s j U Uji. This assignment gives a sheaf. The isomorphisms f i are induced by the projection maps j F j(u U j ) F i (U U i ).

10 10 R. VIRK Now suppose (G, {g i } i I ) is another family satisfying the same properties. Then the isomorphisms g 1 i f i : F Ui G Ui glue to give an isomorphism F G Example. Let U 1, U 2 S 1 be the complements of the north and south pole respectively. Let U ± 12 denote the two connected components of U 1 U 2. Let ζ C. Define a sheaf L ζ on S 1 by gluing C U1 and C U2 as follows. For ε {+, }, let be defined by θ ε : C U1 U ε 12 C U2 U ε 12 θ + = 1 and θ = ζ. If ζ = 1, then we just obtain the constant sheaf C S 1. However, for ζ 1, the sheaf L ζ is not isomorphic to C S Exercise. What is Γ(S 1 ; L ζ )? 12. Additional exercises Exercise. Let f : X Y be a map between spaces. Let G be a presheaf on Y. Construct a canonical isomorphism (f t G) # f 1 (G # ) Exercise. Let X be a space and let i U i = X be an open cover of X. Write j i : U i X for the inclusion map. Let F be a presheaf on X. Suppose j t i F is a sheaf for all i. Prove that F is a sheaf Exercise. Let F : A B and G: B A be functors between categories A and B. Suppose F is left adjoint to G. Show that G preserves limits and F preserves colimits Exercise. Let f : X Y be a map between spaces. In general, is it possible for f to have a right adjoint? Exercise. Let F, G be sheaves on a space X. Show that the assignment U Hom Sh(U) (F U, G U ) defines a sheaf on X (what are the restriction maps?). This sheaf is usually denoted by H om(f, G) and is called sheaf Hom. Note that Γ(X; H om(f, G)) = Hom(F, G) Exercise. Let Define Describe f C X as explicitly as possible. X = {(x, y) R 2 xy 1}. f : X R, (x, y) y Exercise. In this exercise all sheaves are understood to be sheaves of C-vector spaces. Define f : S 1 S 1, z z 2, and g : S 1 S 1, z z 3. The square root sheaf on S 1 is defined to be S = {aφ a C, φ: S 1 S 1 is a continuous map such that fφ = id}. The cube root sheaf on S 1 is defined to be Q = {aφ a C, φ: S 1 S 1 is a continuous map such that gφ = id}.

11 SOME OPERATIONS ON SHEAVES 11 (i) Describe S and Q as explicitly as possible. Remark/hint: See Example Both S and Q are examples of what are called local systems or twisted constant sheaves. We have S x Q x (C S 1) x C. However, no two of these three sheaves are isomorphic to each other. Both S and Q have no global sections. To understand S you may want to pick an open cover of S 1 consisting of two sets such that S restricted to either of these sets is the constant sheaf. Now something interesting will happen with the restriction maps to the intersection of these sets: there will be a twist that will prevent you from gluing sections to obtain a non-trivial global section. Your job is to figure out what this twist is. Now do a similar thing for Q. Can you make an educated guess for the general situation of n-th root sheaves? (ii) Describe the sheaves f C S 1, g C S 1, f S and g Q. (iii) Describe the sheaves f 1 C S 1 and g 1 C S 1. Hint: let h: X Y be any map between spaces. What is h 1 C Y? Exercise. Let U 1, U 2 S 1 be the complements of the north and south poles respectively. Classify (up to isomorphism) all sheaves F (of C-vector spaces) on S 1 satisfying the following property: F U1 C U1 and F U2 C U2. Also describe Γ(S 1 ; F ) for F as above. Now let F 1, F 2 be sheaves on S 1 satisfying the above. Describe Hom(F 1, F 2 ). Hint: Do the previous exercise Exercise. Define a space P 1 as follows. As a set P 1 = C { }. Nontrivial closed sets in P 1 are finite sets. So that non-empty open sets are of the form P 1 {x 1,..., x n }. Define a sheaf O P 1 on P 1 via the assignment U {rational functions U C that have no poles in U}, where a rational function g(x) is said to have no pole at if and only if g( 1 y ) has no pole at y = 0. Compute: (i) Γ(P 1 {0}; O P 1); (ii) Γ(P 1 { }; O P 1); (iii) Γ(P 1 ; O P 1).

12 12 R. VIRK 13. The adjunction (f 1, f ) revisited Let f : X Y be a map of spaces. I will now sketch another proof of the fact that f 1 is left adjoint to f that avoids unit/counit maps Write Sh( ) and PreSh( ) for the category of sheaves and presheaves on, respectively. Let Op( ) denote the category of open sets of Let F Sh(X) and let G Sh(Y ). We need to construct a bifunctorial isomorphism Hom Sh(X) (f 1 G, F ) Hom Sh(Y ) (G, f F ). It suffices to construct a bifunctorial isomorphism Hom PreSh(X) (f t G, F ) Hom PreSh(Y ) (G, f F ). Now an element α Hom PreSh(Y ) (G, f F ) consists of a family of maps {α V : G(V ) F (f 1 (V ))} V Op(Y ) that are compatible with the restriction maps. Equivalently, this is a family of maps {α U : G(V ) F (U)} V Op(Y ),U Op(X),U f 1 (V ) that are compatible with the restriction maps. Consequently, we obtain a family of maps { α U : lim G(V ) F (U)} U Op(X) V f(u) that are compatible with restriction maps. This is precisely the data of an element in Hom PreSh(X) (f t G, F ). A moments thought should convince you that this correspondence gives the require bifunctorial isomorphism. Department of Mathematics, CU-Boulder, Boulder, CO address:

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