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 $U_i$.

  • 1
    $\begingroup$ Why do you think there is something wrong with this argument? $\endgroup$ – Alex Kruckman Apr 30 at 3:07
  • $\begingroup$ 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 $\endgroup$ – 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).


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