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.


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.

  • $\begingroup$ You are right, thanks! $\endgroup$ – Ergos May 1 '15 at 0:12
  • $\begingroup$ I forgot to ask you something very important! What if I defined a morphism of sheaves only on the sections over basis elements. Can I extend this to a morphism to all the open sets? $\endgroup$ – Ergos May 1 '15 at 17:52
  • $\begingroup$ Yes. You might want to have a look at Ravi Vakil's notes. There is a chapter about ' Recovering sheaves from a “sheaf on a base” '. $\endgroup$ – MooS May 2 '15 at 7:27

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