# Why doesn't Hom commute with taking stalks?

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.

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As a supplement to Anton's answer, the identity $Hom(F,G)_p = Hom(F_p,G_p)$ holds in certain general cases. Namely, if $X$ is an algebraic (or complex) variety, the sheaves are sheaves of $\mathcal O_X$-modules, and if $F$ is coherent, then this holds. It's a good exercise to prove this, a precise statement with hints is at the end of Chapter 2 of Demailly's book on complex geometry (available for free on his website). –  Gunnar Magnusson Jan 3 '11 at 14:37
Just to clarify (since this kind of thing confused me a lot in the past), the $Hom$ in @Gunnar's comment above is as $\mathcal O_X$-modules, not as abelian groups. If it were $Hom$ as abelian groups, you could take $F$ to be a coherent skyscraper sheaf and use my first example. –  Anton Geraschenko Jan 3 '11 at 18:27

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.

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Thanks, this was very helpful! –  Vitaly Lorman Jan 3 '11 at 20:00
For the second example, how come $Hom(F,G)(U)$ contains a natural copy of $Hom(H,H)$ for all $U$? Isn't $Hom(F,G)(U) = \mathrm{Hom}(F(U),G(U)) = \mathrm{Hom}(0,0) = 0$ if $p\in U$? –  maxymoo Mar 4 '11 at 4:59
@maxymoo: no, $Hom(F,G)(U)=Hom(F|_U,G|_U)$, which is not the same as $Hom(F(U),G(U))$. –  Anton Geraschenko Mar 4 '11 at 16:09
ah yes, an element of "sheaf hom" i.e. a sheaf morphism is a collection of morphisms, one for each open set. –  maxymoo Mar 4 '11 at 22:10

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.

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Thanks for your help. Instead of Hom, is there a different construction for which both coordinates commute with colimits (perhaps the coproduct)? In this case, could we pull colimits out of each coordinate and expect to get the same thing as taking one colimit for both coordinates? –  Vitaly Lorman Jan 3 '11 at 20:08

Here is a somewhat abstract explanation giving a "deeper reason" for why Hom doesn't commute with taking stalks. In categorical logic, it is well known that in general, only so-called geometric constructions commute with taking inverse image under geometric morphisms. Calculating the stalk at a point is an example of such a process of taking an inverse image.

Anton's answer shows that the Hom construction is not geometric in general.

But there is a very general situation in which $\mathcal{H}om(F, \cdot)$ is geometric. Namely, it suffices for $F$ to be an $\mathcal{O}_X$-module on a ringed space $X$ which is of finite presentation around $x \in X$, i.e. such that there is a short exact sequence $$\mathcal{O}_X^n \longrightarrow \mathcal{O}_X^m \longrightarrow F \longrightarrow 0$$ on an open neighbourhood of $x$. (You can use the constant sheaf $\underline{\mathbb{Z}}$ as the structure sheaf $\mathcal{O}_X$ if you want to stay in the setting of sheaves of abelian groups.) This is because in this case, $\mathcal{H}om(F, G)$ is canonically isomorphic to $$\left\{ x \in G^m \,\middle|\, \sum_i a_{ij} x_i = 0 \in G, j = 1,\ldots,n \right\},$$ where $A = (a_{ij}) \in \mathcal{O}_X^{m \times n}$ is the presentation matrix and $G$ is an arbitrary $\mathcal{O}_X$-module, and this construction is patently geometric.

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