This is exercise 3.24 from Gathmann's notes.
Let $\mathscr{F}$ be a sheaf on a topological space $X$, and let $a\in X$. Show that the stalk $\mathscr{F}_a$ is a local object in the following sense: if $U\subset X$ is an open neighborhood of $a$ then $\mathscr{F}_a$ is isomorphic to the stalk of $\mathscr{F}|_U$ at $a$ on the topological space $U$.
My attempt:
Define the ring isomorphism: $$\overline{(V,\phi)}\mapsto \overline{(V\cap U,\phi|_U)}$$ for any open subset $V\subset X$, that contains $a$.
- It is well-defined. If $(V,\phi)=(V',\phi')$, then $\phi=\phi'$ on some neighborhood of $a\in W\subset V\cap V'$. Thus they agree on $W\cap U$.
- We see that it is surjective trivially, because for any $\overline{(V,\phi)}$ in $U$, $V$ can also be considered as in $X$, together with the function $\phi$ defined on it.
- We show that it is injective. Suppose a function $\phi$ represents the zero function in $\mathscr{O}_{U,a}$, i.e., it is zero in a neighborhood of $a$. Here I am confused. In the case $X$ is not irreducible, can we say that the preimage has to be $0$? Or can we shrink this open neighborhood until it is in $U$?
Thank you for any help!
Also, I am self-learning this subject, sometimes I suspect whether my proof is completely nonsense. If someone could help verify the above proof it would be greatly appreciated!
As requested in the comment, this is the proof in the case of a sheaf of sets.
The map $\overline{(V,\phi)}\mapsto\overline{(V\cap U,\phi_{|V\cap U})}$ is well-defined and surjective. This is done in the original post in the case of a sheaf of rings, but the proof never uses rings and works perfectly for sets.
Let us show that it is also injective. So assume $\overline{(V\cap U,\phi_{|V\cap U})}=\overline{(V'\cap U,\phi'_{|V'\cap U})}$. This mean that there is a neighborhood $W\subset V\cap U$ and $V'\cap U$ such that $(\phi_{|V\cap U})_{|W}=(\phi'_{|V'\cap U})_{|W}$. But this means that $\phi_{|W}=\phi'_{|W}$. So $\overline{(V,\phi)}=\overline{(V',\phi')}$ and the map is indeed injective.