Exactness of sequence of sheaves of abelian group implies exactness of sequence of sections as a group.

Question

How do I prove that exactness of sequence of sheaves of abelian group is equivalent to the sequence of stalk maps at every point?

Or is there any other way I can proceed to prove the title?

Thanks in advance

Answer 1

As you explained in the comments, you wish to deduce the exactness of $0\to F(U)\to G(U)\to H(U)$ from the exactness of $0\to F\to G\to H\to 0$, where $U$ is some open subset of the topological space your sheaves are defined over. In fact, you only require the exactness of $0\to F\to G\to H$. The main point is really that the kernel presheaf is a sheaf. (As this is not true for the cokernel, you will in general not get that $G(U)\to H(U)$ is surjective if $G\to H\to 0$ is exact).

The exactness of the sequence $0\to F(U) \to G(U)\to H(U)$ is saying that the morphism $F(U)\to G(U)$ is the kernel of the latter morphism $G(U)\to H(U)$. By definition of the kernel $K$ of the morphism of sheaves $G\to H$, we have $K(U)=\operatorname{ker}(G(U)\to H(U))$. Since $0\to F\to G\to H$ is exact, the morphism $F\to G$ is the kernel of the morphism $G\to H$. In other words, there exists a canonical isomorphism $F\cong K$ such that the triangle involving the inclusion $K\hookrightarrow G$ (meaning to be the inclusion on sections) and the morphism $F\to G$ commute. Taking the sections over $U$, this will give you a similar commutative triangle. For this just note that a morphism of sheaves is an isomorphism if and only if it is an isomorphism on sections. Hence, the sequence $0\to F(U)\to G(U)\to H(U)$ is exact.