Let $\mathcal F$ and $\mathcal G$ be sheaves of abelian groups over a topological space $X$ and let $\phi : \mathcal F\rightarrow \mathcal G$ be a morphism of sheaves.
If $\{U_{\alpha}\}$ is an open cover of $X$ such that $\phi(U_{\alpha}):\mathcal F(U_{\alpha})\rightarrow\mathcal G(U_{\alpha})$ is an isomorphism of abelian groups for every $\alpha$ then is it true that $\phi$ is an isomorphism of sheaves?
While trying to prove this I had to assume that for an open set $U\subseteq X$ the map $\phi(U_{\alpha}\cap U)$ is an isomorphism. However I can't show this unless the restriction map $\mathcal G(U_{\alpha})\rightarrow\mathcal G(U_{\alpha}\cap U)$ is surjective. So is there another way to prove this or is it false? Can someone give me a counter example?
Thank you.
This is not true. Let $X$ be any topological space, equipped with a nonzero sheaf $\mathcal{F}$ that has no global sections except $0$ (such spaces and sheaves exist). Consider the trivial $1$-element cover of $X$, and let $\phi$ be the $0$-morphism $\mathcal{F}\to\mathcal{F}$. Clearly, the morphism $\phi$ induces an isormorphism $\phi(X):\mathcal{F}(X)\to\mathcal{F}(X)$, but $\phi$ is not an isomorphism of sheaves.
An example of such a sheaf is the sheaf $\mathcal{O}(-1)$ on $\mathbb{P}^1_\mathbb{C}$. Indeed, this is the sheaf of sections of the 'tautological' line bundle, which is a vector bundle whose fiber over a point $p\in\mathbb{P}^1_\mathbb{C}$ corresponding to a line $\ell\subset\mathbb{C}^2$ is the line $\ell$ itself. (The specifics of the proof here depend a bit on what category you're working in, algebraic, analytic, topological...).