Let $X$ be a topological space with sheaves of abelian groups $\mathcal{F}$ and $\mathcal{G}$. I want to know under what conditions I can say that these sheaves are isomorphic. By "isomorphic", I mean there exists an isomorphism between them, and by "isomorphism", I mean a morphism of sheaves with a two-sided inverse.
I know it is not sufficient that they agree on stalks. However, if a morphism of sheaves induces an isomorphism on stalks, then they are isomorphic.
I also know that if a morphism of sheaves has trivial kernel and cokernel, then it is an isomorphism.
I was previously under the impression that if you have a base $\mathfrak{B}$ for the topological space, and if $\mathcal{F}(U) \simeq \mathcal{G}(U)$ for every $U \in \mathfrak{B}$, then $\mathcal{F}$ and $\mathcal{G}$ are isomorphic. However, I was informed recently that is not the case. Does anyone have a counter example for this?
So now I am beginning to wonder, if $\mathcal{F}(V) \simeq \mathcal{G}(V)$ for every open $V \subseteq X$, can we even say that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic then, or not? If not, does someone have a counter example?
Each manifold $M$ has a base $\frak B$ consisting of $U$ all homeomorphic to $\Bbb R^n$. If $\mathcal{F}$ is the sheaf of sections of a rank $r$ vector bundle, then $\mathcal{F}(U)$ are isomorphic to the group of continuous maps from $U$ to $\Bbb R^n$. If we have two such bundles, then the sheaves satisfy $\mathcal{F}(U) \cong\mathcal{G}(U)$ for all $U\in \frak B$, but not all vector bundles of the same rank are isomorphic.
Come to think of it one can do better. I think this works. Consider a projective curve $C$ in the Zariski topology. Let $\mathcal{F}$ and $\mathcal{G}$ be the sheaves corresponding to line bundles on $C$. Then for proper open $U$, $\mathcal{F}(U)$ and $\mathcal{G}(U)$ are free rank-one modules over $\mathcal{O}(U)$ where $U$ is the structure sheaf. One can arrange $\mathcal F(C)$ and $\mathcal G(C)$ to be both zero and $\mathcal F$ and $\mathcal G$ to be non-isomorphic bundles. To do this, take these to be bundles associated to divisors $P$ and $2P$, where $P$ is a point, where the sections of the bundle associated to a divisor $D$ consists of the functions with no poles and zeroes of the right order at the points of $D$.
Sorry, this is all a bit vague, as I haven't the time to check the right notation in say Hartshorne's book, but I contend there are very natural examples from algebraic geometry of non-isomorphic sheaves with all groups of sections over open sets isomorphic.