# In which cases does pullback commute with the Hom-sheaf?

Assume $f: (X,\mathcal{O}_X)\rightarrow (Y,\mathcal{O}_Y)$ is a morphism of locally ringed spaces and E and F are two locally free $\mathcal{O}_Y$-moduels of finite rank.

I was wondering if we have the following isomorphism in this case: $f^{*}\mathcal{H}om_Y(E,F)\cong \mathcal{H}om_X(f^{*}E,f^{*}F)$?

My idea was the following: let $G$ be an $\mathcal{O}_X$-module, then we have $Hom(\mathcal{H}om_X(f^{*}E,f^{*}F),G)\cong Hom( f^{*}F\otimes (f^{*}E)^\vee,G)\cong Hom(f^{*}F,\mathcal{H}om((f^{*}E)^{\vee},G))\cong Hom(f^{*}F,f^{*}E\otimes G)\cong Hom(F,f_{*}(f^{*}E\otimes G))\cong Hom(F,E\otimes f_{*}G)\cong Hom(\mathcal{H}om_Y(E,F),f_{*}G)\cong Hom(f^{*}(\mathcal{H}om_Y(E,F)),G)$

I used that E is locally free of finite rank => $f^{*}E$ is locally free of finite rank => $\mathcal{H}om(f^{*}E,f^{*}F)\cong (f^{*}E)^{\vee}\otimes f^{*}F$

$f^{*}E$ is locally free of finite rank => $(f^{*}E)^{\vee\vee}\cong f^{*}E$

$E$ is locally free of finite rank => $f_{*}(f^{*}E\otimes G)\cong E\otimes f_{*}G$ (projection formula)

So Yoneda should give us $\mathcal{H}om_X(f^{*}E,f^{*}F)\cong f^{*}\mathcal{H}om_Y(E,F)$?

Is my idea correct? If so, does this isomorphism also hold in other cases? For example is suspect that it could be true for $E$ a coherent sheaf and $f$ a flat morphism, where it would boil down to the known fact that from commutative algebra

$Hom_A(M,N)\otimes B \cong Hom_B(M\otimes B, N\otimes B)$

on the stalk level and $B$ a flat $A$-module, if we had a map $f^{*}\mathcal{H}om(E,F)\rightarrow \mathcal{H}om(f^{*}E,f^{*}F)$. But do we alsways have a sheaf morphism between these two sheaves on $X$? I cannot seem to construct such a map.

• A belated question. How does the penultimate isomorphism in your long chains of isomorphisms between hom sets follow? Tensor-Hom adjunction seems to go the other way. I was able to follow the other ones, but not this one. Thanks in advance! Commented Jan 27, 2018 at 20:46
• @LeeWang you can replace tensoring with a locally free sheaf in one of the argumens by tensoring with its dual in the other argument and you can use the isomorphism from the tensor product of the dual of a vector space times another vector space to the linear maps from the first vector space to the second. That you can do this to move something from the first argument to the second follows from tensor hom adjunction. The other way around follows from previous case and that the double dual of a locally free sheaf is itself Commented Aug 12, 2018 at 18:55

## 1 Answer

$\def\H{{\mathcal Hom}}\def\HH{{\operatorname{Hom}}}$Everything you wrote seems correct for the locally free case. Though its easy to overlook subtle things with these type of arguments, it looks good to me.

For the second I believe there is a natural map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$. I will actually define a map $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$ and use the adjunction with $f^*$ to get the required map.

Let $U \subseteq Y$ be open. Then $\H_Y(E,F)(U) = \HH(E|_U,F|_U)$ and $f_*\H_X(f^*E,F^*F)(U) = \HH(f^*E|_{f^{-1}(U)}, f^*F|_{f^{-1}(U)})$. Then we note that

$$f^*E|_{f^{-1}(U)} = f|_{f^{-1}(U)}^*\left(E|_{f^{-1}(U)}\right)$$

and $f^*|_{f^{-1}(U)}$ is a functor from sheaves of $\mathcal{O}_U$ modules to $\mathcal{O}_{f^{-1}(U)}$ modules. Thus we get a natural map of sheaves of modules from $U$ from functoriality:

$$\HH(E|_U, F|_U) \to \HH\left(f|_{f^{-1}(U)}^*\left(E|_{f^{-1}(U)}\right), f|_{f^{-1}(U)}^*\left(F|_{f^{-1}(U)}\right)\right) = \HH(f^*E|_{f^{-1}(U)}, f^*F|_{f^{-1}(U)}).$$

Everything is suitably natural enough that it should commute with the restriction maps for inclusions $V \subset U$ giving a map of sheaves $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$.

Now that we have the required map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$, I think we need $X$ and $Y$ to be schemes not just locally ringed spaces. Then we can reduce to checking that this is an isomorphism on affine covers in which case it reduces to the isomorphism from commutative algebra. Without having $X$ and $Y$ be schemes I don't think we can make the argument work because $\H$ does not commute with taking stalks so we can't just check it on local rings.

EDIT: See comments below, apparently the argument does work on any ringed space as long as $E$ is of finite presentation.

• Thanks. Yes, you are right i think we need schemes for the stalk argument. But since i only work with schemes this is fine for me. I only mentioned locally ringed spaces, because usually one can prove such results in a more general framework.
– DonD
Commented Jan 21, 2015 at 9:41
• Possibly $X,Y$ need not have to be schemes. We have the following result: Let $(X, \mathcal O_X)$ be a ringed space and let $\mathcal F$ be an $\mathcal O_X$-module of finite presentation. Then for each $x \in X$ and for each $\mathcal O_X$-module $\mathcal G,$ the canonical homomorphism of $\mathcal O_{X,x}$-modules $\mathcal{H}om_{\mathcal O_{X}}(\mathcal F, \mathcal G)_x \to \text{Hom}_{\mathcal O_{X,x}}(\mathcal F_x, \mathcal G_x)$ is bijective. (For a proof see U. Gortz, T. Wedhorn: Algebraic Geometry 1, Chapter 7, Prop: 7.27.) Commented Jan 21, 2015 at 11:05
• @Krish Oh very nice. After writing my answer I found this result stated for coherent sheaves on a scheme but I hadn't seen it for general ringed spaces. Thanks! Commented Jan 21, 2015 at 11:13
• @Krish : Thanks. This is a nice book. Exercise 7.20. deals exactly with my problem! It says that the map is an isomorphism if E is locally free of finite rank or if f is flat und E is of finite presentation. And it works on ringed spaces!
– DonD
Commented Jan 21, 2015 at 11:15
• Both of you are most welcome. Commented Jan 21, 2015 at 12:18