This is Exercise 2.4.E. from Vakil's notes.
Show that a morphism of shaves of sets is an isomorphism if and only if it induces an isomorphism of all stalks.
My failed attempts:
I found some similar questions here and here. But I still couldn't figure out the details.
I have shown the "only if" part. I have also shown that if the morphism induces an isomorphism of stalks, it is injective.
The surjective part confused me a lot. Here is a commutative diagram, where the vertical maps are injections:
$$\require{AMScd}
\begin{CD}
\mathscr{F}(U) @>{\phi(U)}>> \mathscr{G}(U)\\
@VVV @VVV \\
\prod_{p\in U} \mathscr{F}_p @>{\phi_p}>> \prod_{p\in U} \mathscr{G}_p
\end{CD}$$
To show surjectivity, suppose $g\in \mathscr{G}(U)$. Its image in the stalks at $p$ is $(g,U)_p$. Since $\phi_p$ is an isomorphism for all $p$, we have $(f_p,U_p)\in \mathscr{F}_p$, such that $\phi_p(f_p,U_p)=(g,U)_p$ for all $p\in U$. Now I don't know how to glue the $f_p$ together. Since even we have the pairs $(f_q,U_q)\in \mathscr{F}_q$, we cannot show that their restrictions on $U_p\cap U_q$ are equal. We have to find a small enough open set to glue them.
I also tried the open cover argument from the right hand side:
Since $\prod(g,U)_p$ consists of compatible germs, there exists an open cover $U_i$, and section $g_i$ in $\mathscr{G}(U_i)$, such that $(g,U)_p=(g_i,U_i)$ for all $p\in U_i$. But this doesn't help much. Since $\phi_p$ are different maps, we still cannot conclude anything about the $(f_p,U_p)$.
Any help will be greatly appreciated!
Best Answer
Look carefully at the definition of surjectivity of a morphism of sheaves.
It is not required that $\mathcal{F}(U)\to\mathcal{G}(U)$ is surjective for all $U$, only that, for each $g\in \mathcal{G}(U)$, there is some cover $\{U_i\}$ of $U$ such that $g|_{U_i}$ is in the image of $\mathcal{F}(U_i)\to\mathcal{G}(U_i)$ for all $i$.
In fact, you have already shown this.
I just remembered that surjectivity of $\mathcal{F}(U)\to\mathcal{G}(U)$ follows from the above condition, plus injectivity.
To see this, suppose that $f_i\mapsto g_i$. Then $f_i |_{U_i\cap U_j} \mapsto g_i |_{U_i\cap U_j}$ and $f_j |_{U_i\cap U_j} \mapsto g_j |_{U_i\cap U_j}$. But since $g_i |_{U_i\cap U_j} = g_j |_{U_i\cap U_j}$, we have $f_i |_{U_i\cap U_j} = f_j |_{U_i\cap U_j}$ by injectivity, so we can glue the $f_i$ to an $f$ with $f\mapsto g$.