
## 17.20 Internal Hom

Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. Consider the rule

$U \longmapsto \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X|_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U).$

It follows from the discussion in Sheaves, Section 6.33 that this is a sheaf of abelian groups. In addition, given an element $\varphi \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X|_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U)$ and a section $f \in \mathcal{O}_ X(U)$ then we can define $f\varphi \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X|_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U)$ by either precomposing with multiplication by $f$ on $\mathcal{F}|_ U$ or postcomposing with multiplication by $f$ on $\mathcal{G}|_ U$ (it gives the same result). Hence we in fact get a sheaf of $\mathcal{O}_ X$-modules. We will denote this sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$. There is a canonical “evaluation” morphism

$\mathcal{F} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathcal{G}.$

For every $x \in X$ there is also a canonical morphism

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \mathcal{G}_ x)$

which is rarely an isomorphism.

Lemma 17.20.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ be $\mathcal{O}_ X$-modules. There is a canonical isomorphism

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X} (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}, \mathcal{H}) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X} (\mathcal{F}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{H}))$

which is functorial in all three entries (sheaf Hom in all three spots). In particular, to give a morphism $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{H}$ is the same as giving a morphism $\mathcal{F} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{H})$.

Proof. This is the analogue of Algebra, Lemma 10.11.8. The proof is the same, and is omitted. $\square$

Lemma 17.20.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules.

1. If $\mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F} \to 0$ is an exact sequence of $\mathcal{O}_ X$-modules, then

$0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}_1, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}_2, \mathcal{G})$

is exact.

2. If $0 \to \mathcal{G} \to \mathcal{G}_1 \to \mathcal{G}_2$ is an exact sequence of $\mathcal{O}_ X$-modules, then

$0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_1) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_2)$

is exact.

Proof. Omitted. $\square$

Lemma 17.20.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is finitely presented then the canonical map

$\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \mathcal{G}_ x)$

is an isomorphism.

Proof. By localizing on $X$ we may assume that $\mathcal{F}$ has a presentation

$\bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{O}_ X \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ X \to \mathcal{F} \to 0.$

By Lemma 17.20.2 this gives an exact sequence $0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G} \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}.$ Taking stalks we get an exact sequence $0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G}_ x \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}_ x$ and the result follows since $\mathcal{F}_ x$ sits in an exact sequence $\bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{O}_{X, x} \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_{X, x} \to \mathcal{F}_ x \to 0$ which induces the exact sequence $0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{F}_ x, \mathcal{G}_ x) \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G}_ x \longrightarrow \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}_ x$ which is the same as the one above. $\square$

Lemma 17.20.4. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ Y$-modules. If $\mathcal{F}$ is finitely presented and $f$ is flat, then the canonical map

$f^*\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f^*\mathcal{F}, f^*\mathcal{G})$

is an isomorphism.

Proof. Note that $f^*\mathcal{F}$ is also finitely presented (Lemma 17.11.4). Let $x \in X$ map to $y \in Y$. Looking at the stalks at $x$ we get an isomorphism by Lemma 17.20.3 and More on Algebra, Remark 15.62.21 to see that in this case $\mathop{\mathrm{Hom}}\nolimits$ commutes with base change by $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$. Second proof: use the exact same argument as given in the proof of Lemma 17.20.3. $\square$

Lemma 17.20.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is finitely presented then the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is locally a kernel of a map between finite direct sums of copies of $\mathcal{G}$. In particular, if $\mathcal{G}$ is coherent then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is coherent too.

Proof. The first assertion we saw in the proof of Lemma 17.20.3. And the result for coherent sheaves then follows from Lemma 17.12.4. $\square$

Lemma 17.20.6. Let $X$ be a topological space. Let $\mathcal{O}_1 \to \mathcal{O}_2$ be a homomorphism of sheaves of rings. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_1}(\mathcal{F}_{\mathcal{O}_1}, \mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_2}(\mathcal{F}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_1}(\mathcal{O}_2, \mathcal{G}))$

bifunctorially in $\mathcal{F} \in \textit{Mod}(\mathcal{O}_2)$ and $\mathcal{G} \in \textit{Mod}(\mathcal{O}_1)$.

Proof. Omitted. This is the analogue of Algebra, Lemma 10.13.4 and is proved in exactly the same way. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).