morphism betweem sheaves that is an isomorphism in local sections of a basis

Question

Let $\{U_{\alpha}:\alpha \in A\}$ be a basis of open sets for the topological space $X$.

Let $\mathscr{F},\mathscr{G}$ be sheaves over $X$. Suppose that there exist a morphism $\phi: \mathscr{F} \to \mathscr{F}$ such that:

$$\phi_U:\mathscr{F}(\{U_{\alpha}\}\to \mathscr{G}(\{U_{\alpha}\} $$

is an isomorphism for each $\alpha \in A$.

It's true that the morphism $\phi$ is in fact an isomorphism?.

I was working over locally ringed spaces and I constructed an morphism betweem them. This morphism is in fact an isomorphism on the local sections defined on the basis elements.

In my case the ringed spaces are in fact isomorphic. I read other proof and I know that my morphism is in fact an isomorphism, but only because I read other proof.

Answer 1

Your assumptions immediately imply, that your morphism is an isomorphism on each stalk, since any point lies in some basis open set and passing to a stalk is functorial.

So the answer is yes.