Constructing Associated Sheaf of Abelian Groups

[disregardthat]

Given a presheaf F of abelian groups, we want to construct an associated sheaf G and a morphism f from F to G such that given a morphism h from F to a sheaf H, there is a unique morphism g from G to H such that h = gf. Furthermore, G will be unique up to isomorphism.

We construct it by letting the abelian group for an open subset U be G(U) = the set of functions $$s : U \to \bigcup_{P \in U} F_P$$, where $$F_P$$ denotes the stalk of F at P satisfying the following criterion:

1) $$s(P) \in F_P$$ for all P in U, and
2) For any P in U, there is a neighbourhood V of P in U such that for any Q in V, there is a t in F(V) such that the stalk of t in $$F_Q$$ is equal to s(Q).

I have proved that G is a sheaf, and I think we can define f by defining f(U) : F(U) --> G(U) by letting f(U)(s) be the constant function of (s,U), a germ in $$F_P$$ for all P.

Now, given an h from F to H, how can I define g : G --> H? The book says it's natural, but I just don't see it.

I would also appreciate some hints as to how to prove uniqueness of g after knowing how it is defined. I have proven uniqueness of G given uniqueness of g.

How can we show that $$F_P = G_P$$ for all P?

Really appreciate any help or comments!

 

[micromass]

Given a presheaf F of abelian groups, we want to construct an associated sheaf G and a morphism f from F to G such that given a morphism h from F to a sheaf H, there is a unique morphism g from G to H such that h = gf. Furthermore, G will be unique up to isomorphism.

We construct it by letting the abelian group for an open subset U be G(U) = the set of functions $$s : U \to \bigcup_{P \in U} F_P$$

don't you mean the disjoint union here? Or did you construct the stalks such that they are disjoint?

, where $$F_P$$ denotes the stalk of F at P satisfying the following criterion:

1) $$s(P) \in F_P$$ for all P in U, and
2) For any P in U, there is a neighbourhood V of P in U such that for any Q in V, there is a t in F(V) such that the stalk of t in $$F_Q$$ is equal to s(Q).

I have proved that G is a sheaf, and I think we can define f by defining f(U) : F(U) --> G(U) by letting f(U)(s) be the constant function of (s,U), a germ in $$F_P$$ for all P.

Now, given an h from F to H, how can I define g : G --> H? The book says it's natural, but I just don't see it.

Try this: the morphism f determines a morphisms on the stalks. Thus for P, we have a morphism $$f_P:F_P\rightarrow H_P:s\rightarrow (f(s))_P$$. Taking all these morphisms will get you a morphism $$\bigcup_{P}{F_P}\rightarrow \bigcup_{P}{H_P}$$. Composing with this morphism will get you the desired morphism of sheaves. So the only thing you still need to do is show that $$\bigcup_{P}{H_P}$$ is a sheaf isomorphic to H.

I would also appreciate some hints as to how to prove uniqueness of g after knowing how it is defined. I have proven uniqueness of G given uniqueness of g.

The stalks of g determine g completely. So you must prove that g is unique in the stalks of the sheaf.

How can we show that $$F_P = G_P$$ for all P?

I suggest you show this first. You'll need to show that F_P is the direct limit of the G(U)'s. Have you put a topology on $$\bigcup_{P}{H_P}$$ yet. You'll have to use that the map $$p:\bigcup_{P}{H_P}\rightarrow X$$ (with $$p^{-1}(P)=H_P$$) is a local homeomorphism...

May I ask you from what book you're learning sheaves. It's pretty hard to explain without knowing which notations you're familiar to...

 

[disregardthat]

Thank you for your reply micromass, but I have figured my questions out now. I misinterpreted the definition of the functions s, the sections of G. I thought initially that for each Q in V there is a t depending on Q in F(V) such that s(Q) = the stalk of t. It was the other way around, t is not depending on Q but only on V. It all went much more smoothly after I realized that.

By $$\bigcup F_P$$ I meant union, the author (Robin Hartshorne on Algebraic Geometry) presumably means the set of germs (V,s) which are equal if they are equal on some common open subset by the restriction morphisms. The classes agree on intersections of stalks, so we can assume they are all part of the set of all germs. Or we can just consider them all subsets of the direct limit of the abelian groups corresponding to every open subset.

EDIT: A topology on $$\bigcup H_P$$ never came in question, note that the union is only over P in the open subset U, not all of X. That F_P = G_P was actually quite easy, one can just choose a sufficiently small open subset of P in the domain of the stalks of functions in G on which they are constant, and we immediately have a 1-1 correspondence induced by the map from F to G.

EDIT again: I think there have been a slight miscommunication. The functions $$s : U \to \bigcup_{P \in U} F_P$$ satisfying 1) and 2) are supposed to make up the abelian group G(U).