Topics in Abstract Algebra/Sheaf theory
We say is a pre-sheaf on a topological space if
- (i) is an abelian group for every open subset
- (ii) For each inclusion , we have the group morphism such that
- is the identity and for any inclusion
A pre-sheaf is called a sheaf if the following "gluing axiom" holds:
- For each open subset and its open cover , if are such that in , then there exists a unique such that for all .
Note that the uniqueness implies that if and for all , then . In particular, for all implies .
4 Example: Let be a topological group (e.g., ). Let be the set of all continuous maps from open subsets to . Then forms a sheaf. In particular, suppose the topology for is discrete. Then is called a constant sheaf.
Given sheaves and , a sheaf morphism is a collection of group morphisms satisfying: for every open subset ,
where the first is one that comes with and the second .
Define for each open subset . is then a sheaf. In fact, suppose . Then there is such that . But since
for all , we have . Unfortunately, does not turn out to be a sheaf if it is defined in the same way. We thus define to be the set of all such that there is an open cover of such that is in the image of . This is a sheaf. In fact, as before, let be such that . Then we have an open cover of such that restricted to each member of the cover is in the image of .
Let be sheaves on the same topological space.
A sheaf on is said to be flabby if is surjective. Let , and, for each , define . is closed since implies has a neighborhood of such that for every . Define . In particular, if is a closed subset and , then the natural map is an isomorphism.
4 Theorem Suppose
is exact. Then, for every open subset
is exact. Furthermore, is surjective if is flabby.
Proof: That the kernel of is trivial means that has trivial kernel for any . Thus the first map is clear. Next, denoting by , suppose with . Then there exists an open cover of and such that . Since in and is injective by the early part of the proof, we have in and so we get such that . Finally, to show that the last map is surjective, let , and . If is totally ordered, then let . Since agree on overlaps by totally ordered-ness, there is with . Thus, is an upper bound of the collection . By Zorn's Lemma, we then find a maximal element . We claim . Suppose not. Then there exists with . Since in , by the early part of the proof, there exists with . Then (so ) while in . This contradicts the maximality of . Hence, we conclude and so .
is exact if and only if
is exact for every .
Suppose is a continuous map. The sheaf (called the pushforward of by ) is defined by for an open subset . Suppose is a continuous map. The sheaf is then defined by the sheafification of the presheaf where is an open subset of . The two are related in the following way. Let be an open subset. Then consists of elements in where . Since , we find a map
by sending to . The map is well-defined for it doesn't depend on the choice of .