Algebraic Analysis notes Lecture 6 (18 Jan 2019)

Notes for Lecture 5

Last time, for a given topological space X and a ring k, we define a presheaf to be a functor F \colon Op(X)^{op} \to k-mod.

Definition: If F is a presheaf on X, and x is a point in X, then the stalk of F at x, is the k-module F_x = \varinjlim_{x \in U} F(U).

Let’s unpack this: an element of F_x is an equivalence class of pairs (U, s) where U is an open neighborhood of x, and s \in F(U), and (U, s) \sim (U', s') if there is a set V contained in U \cap U' such that s|_V = s'|_V, a “common refinement”.

If x \in U, there is a canonical map F(U) \to F_x, and the image of a section s \in F(U) under this map is called the germ of s at x, denoted s_x.

Definition: Let F \in PreSh(X), the support of F is supp = \overline{ \{ x \in X | F_x \neq 0 \} }. For s \in F(U), let supp(s) = \{x \in U | s_x \neq 0 \}.

Exercise: supp(s) is always closed (that’s why the definition doesn’t have to include taking the closure).

Definition: A presheaf F on X is a sheaf if it satisfies the following:
(S1) Gluing: given an open cover (U_\alpha)_{\alpha \in I} of U open in X, and s_\alpha \in F(U_\alpha) such that s_\alpha |_{U_\alpha \cap U_\beta} = s_\beta |_{U_\alpha \cap U_\beta}, then there exists s \in F(U) such that s|_{U_\alpha} = s_\alpha.
(S2) Local Identity: given an open cover (U_\alpha)_{\alpha \in I} of U open in X, and s,t \in F(U) such that s|_{U_\alpha} = t|_{U_\alpha}, then s=t.

We’re thinking of F(U) as the set of functions from U into k. So the first condition (S1) is saying that if we have a cover of U by open sets, and we have a function on each of these opens, and they all agree on overlaps, then we can glue them all together into a function on all of U. The second condition (S2) says that if two sections agree for any restriction to a set in the open cover, then they must be equal. Taken together, if you have two candidates for what the gluing of a compatible family of sections s_\alpha could be, they must be equal, i.e. they glue to a unique section s on U.

A morphism of sheaves is just a morphism of presheaves. So the category of sheaves on X is the full subcategory of PreSh(X).

Notice that if F is a sheaf, then the local identity condition gives F(\emptyset) = 0. If F is a sheaf, and (U_\alpha)_{\alpha \in I} is an oper cover of U, we get a sequence

where r_0 is given by s \mapsto (s|_{U_\alpha})_{\alpha \in I}, and r_1 is given by (s_\alpha)_{\alpha \in I} \mapsto (s_\alpha|_{U_\alpha \cap U_\beta} - s_\beta|_{U_\alpha \cap U_\beta}). F being a presheaf gives that r_1 \circ r_0 = 0. The gluing condition gives that ker(r_1) \subset Im(r_0). The local identity condition gives that r_0 is injective. Thus if F is a sheaf, then this sequence is exact.

Lemma: Let F \in PreSh(X). The following are equivalent:

  1. F satisfies local identity (S2)
  2. for any U open in X, and s \in F(U), if s_x = 0 for any point x in U, then s = 0.
  3. for any U open in X, and s,t \in F(U), if s_x = t_x for each point x in U, then s = t.

proof: (1 implies 2) Suppose s \in F(U) and s_x = 0 for each x in U. Then for each x in U, there is a neighborhood U^{(x)} of x such that s|_{U^{(x)}} = 0. So we have an open cover (U^{(x)})_{x \in U} of U, and s|_{U^{(x)}} = 0, so local identity gives s=0.

(2 implies 3) Suppose we have s,t \in F(U) with s_x = t_x for each point x in U. Then if we consider s-t, it reduces to the statement of 2.

(3 implies 1) Suppose (S2) fails. Then there is some U open in X, and some s \neq t \in F(U), and some open cover (U_\alpha)_{\alpha \in I} such that s|_{U_\alpha} = t|_{U_\alpha} for every \alpha \in I. Then the germs of s and t agree, s_x = t_x, and thus s=t by 3. Contradiction!

Conclusion: any sheaf can be recovered from the germs of its sections!

Example: First a non-example. The constant presheaf \underline M_{pre,X} is usually not a sheaf because \underline M_{pre,X} (\emptyset) = M which is required to be trivial.

Definition: The constant sheaf, denoted \underline M_X or \underline M when the space is clear, is given by \underline M(U) = \{\text{locally const functions } U \to M\}, and res_{U,V} is just given by restriction of functions.

More generally, for all Z \subset X closed, we write \underline M_{Z \subset X} or just \underline M_Z for \underline M_{Z}(U) = \{ \text{loc const fun.s } U \cap Z \to M\}.

For example, if X is Hausdorff, then for a point x in X, {x} is a closed subset, so then \underline M_x(U) is M if x is in U, and 0 otherwise.

Next time: Sheafification.

Notes for Lecture 7

Published by Joe Moeller

Mathematician

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