# sheaf isomorphisms does not necessarily glue to be isomorphism

For an isomorphism of sheaf on X $$f : \mathscr{F} \to \mathscr{G}$$, suppose it is "locally isomorphic," that is, there is an open cover $$U_i$$ such that $$f|U_i: \mathscr{F}_{U_i} \to \mathscr{G}_{U_i}$$ is an isomorphism for all $$U_i$$.

I cannot specify where is wrong about the argument below :

$$f$$ is isomorphic

iff $$f_x$$ is isomorphic for all $$x\in X$$

iff $$f_x$$ is isomorphic for all $$U_i$$ and $$x\in U_i$$

iff $$f|U_i$$ is isomorphic for all $$U_i$$.

• Why do you think there is something wrong with this argument? – Alex Kruckman Apr 30 at 3:07
• I cosidered problem "locally open embedding morphisms are not necessarily open embeddings", and I tried to generalize the situation. But it seems that I went wrong – Angol Mois Apr 30 at 3:46

Your equivalences you're listing are correct. The isomorphisms already glue since they given by $$f$$. For example, the correct statement for the second point is "There exist a morphism $$f : \mathscr F \to \mathscr G$$ such that for all $$x \in X$$, $$f_x$$ is an isomorphism."
The wrong statement would be to assume that if $$f_x : \mathscr F_x \to \mathscr G_x$$ is an arbitrary family family of isomorphisms, then $$\mathscr F \cong \mathscr G$$ are isomorphic (e.g a counter-example is given by any non-trivial locally free sheaf).