# A morphism of sheaves of sets is an isomorphism if it induces an isomorphism of all stalks.

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:

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!

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 $f\mapsto g$.

• Thank you very much! That clarifies a lot! But I don't think I have seen this definition before. In fact the definition of morphism of presheavs is defined briefly and nothing has been mentioned about injectivity or surjectivity in the notes. – KittyL Jan 5 '17 at 20:16
• Just to clarify, the proof doesn't even use the compatiblibity of $g$, right? Since the $U_p$ already construct an open cover. – KittyL Jan 5 '17 at 20:25
• @KittyL One reason for this definition of surjectivity is that the image presheaf is not a sheaf: you must take the sheafification of the image presheaf (see Exercise 2.5.C). See also Exercise 2.7.E. – André 3000 Jan 6 '17 at 0:23
• @KittyL We need each $g_i$ to be the restriction of $g$, so we need compatibility in that sense. As for your first comment, this looks like it might be an omission in Vakil's notes. Maybe we can fix the terminology problem by remembering that "isomorphism" should really mean "morphism with a two-sided inverse morphism", not "morphism that is injective and surjective". (As an exercise, prove that a morphism of sheaves is an isomorphism if and only if it is injective and—in the sense I gave in my answer—surjective. Maybe this is what Vakil had in mind for this exercise.) – Slade Jan 6 '17 at 1:09
• @SpamIAm: I think I see what you mean now. I didn't consider the image as a presheaf. This is interesting connection. I'm going to read the next section which is about sheafification. I'll understand more about this after that. Thank you! – KittyL Jan 6 '17 at 18:20