Why doesn't Hom commute with taking stalks?

Question

I have been learning about sheaves and am thinking about the following problem. Let $F$ and $G$ be sheaves, say of abelian groups, on a space $X$. The sheaf $Hom(F, G)$ is defined by $Hom(F, G)(U)=Mor(F|_U, G|_U)$. Given a point $p \in X$ and an open set $U$ containing $p$, a morphism $\varphi: F|_U \rightarrow G|_U$ induces a homomorphism on stalks $\phi: F_p \rightarrow G_p$, which is an element of $Hom(F_p, G_p)$. Thus, by the universal property of direct limits, we have a homomorphism from $Hom(F, G)_p$ to $Hom(F_p, G_p)$. However, this is not in general injective or surjective. Why not? An example or a hint leading towards an example would be much appreciated. I have thought about this for some simple sheaves (such as skyscraper sheaves), but it seems to be true in those cases.

I am also interested in a more general answer if there is one, i.e. something category theoretic about Hom and direct limits.

Answer 1

Here's an answer to your call for examples. Take $p$ to be a non-isolated closed point in your favorite topological space $X$. Let $H$ be a non-trivial abelian group.

$Hom(F,G)_p\to Hom(F_p,G_p)$ fails to be surjective

Let $G$ be the constant sheaf $H$, and let $F$ be the skyscraper sheaf at $p$ with stalk $H$. Then $Hom(F,G)$ is the zero sheaf: any section of $F$ is trivial away from $p$, so a homomorphism would take it to a section of $G$ which is trivial away from $p$, but any section of $G$ which is trivial away from $p$ must be trivial, so every section of $F$ must be taken to the trivial section of $G$. So $Hom(F,G)_p=0$, but $Hom(F_p,G_p)=Hom(H,H)\neq 0$.

$Hom(F,G)_p\to Hom(F_p,G_p)$ fails to be injective

Let $V=X\setminus p$, and let $F=G$ be the extension by zero of the constant sheaf $H$ on $V$ (i.e. it's the constant sheaf $H$ on $V$ and $F(U)=0$ if $U\not\subseteq V$). Then $F_p=G_p=0$, but $Hom(F,G)(U)$ contains a natural copy of $Hom(H,H)\neq 0$ for any $U$, so $Hom(F,G)_p$ contains a natural copy of $Hom(H,H)\neq 0$, so it's not zero.

Answer 2

This bad behavior is due to the righthand coordinate of Hom not commuting with colimits (direct limits are a special type of filtered colimit, so this addresses your question), and the lefthand coordinate flips them into limits (this is a general phenomenon in category theory and follows from the definition of the limit (resp. colimit)).

I should also note that you're trying to turn two colimits into one, and I don't really see why you would expect this to work.

For instance, the following deduction is valid: $Hom(F_p,G_p)\cong Hom(colim_{p\in U}F(U),G_p)\cong lim_{p\in U}Hom(F(U),G_p)$. However, this looks like a very different mathematical object than the one you mentioned. Assuming, for the sake of argument, that $F(U)$ is finitely presented for all $U$, we can pull the colimit out front as well, but this doesn't really get us any closer to our stated goal.