Let $X$ be a topological space and let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over $X$.


Of course, if one has a morphism $f : \mathcal{F} \to \mathcal{G}$ such that for all $x\in X$, $f\_x : \mathcal{F}\_x \to \mathcal{G}\_x$ is an isomorphism, then it is known that $f$ itself is an isomorphism.


My question is the following: if we don't have such a morphism $f$, but if we know that for all $x\in X$, $\mathcal{F}\_x$ and $\mathcal{G}\_x$ are isomorphic, is it true that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic ?



Are two sheaves that are locally isomorphic globally isomorphic ?
Definitely not. If $X$ is a ringed space, then a $\mathcal{O}\_X$-module $F$ is called locally free of rank $1$, if $X$ is covered by open subsets $U\_i$ such that $F|\_{U\_i}$ is free of rank $1$ over $\mathcal{O}\_{U\_i}$. The correspond to line bundles on $X$. Line bundles form a group, called the Picard group of $X$ and denoted by $\text{Pic}(X)$. This group does not have to vanish: If $X$ is a CW complex and we take the sheaf of continuous functions, then $\text{Pic}(X)$ is isomorphic to $H^1(X,\mathbb{Z}/2)$. For $X=S^1$, the moebius strip is the nontrivial element here. The corresponding example in algebraic geometry is $\text{Pic}(\mathbb{P}^n\_k)=\mathbb{Z}$, the generator given by the Serre twist $\mathcal{O}(1)$.


Of course, there are more easy counterexamples, but I wanted to indicate that there is a rich theory coming from the observation that two locally isomorphic sheaves are not isomorphic.


