17.22 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 17.22.1. 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 10.12.8. The proof is the same, and is omitted.
$\square$
Lemma 17.22.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules.
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.
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.
Let $\mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F} \to 0$ be as in (1). For every $U \subset X$ open the sequence
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}_1|_ U, \mathcal{G}|_ U) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}_2|_ U, \mathcal{G}|_ U) \]
is exact by Homology, Lemma 12.5.8. This means that taking sections over $U$ of the sequence of sheaves in (1) produces an exact sequence of abelian groups. Hence the sequence in (1) is exact by definition. The proof of (2) is exactly the same.
$\square$
Lemma 17.22.3. 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 10.14.4 and is proved in exactly the same way.
$\square$
Lemma 17.22.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. If $\mathcal{F}$ is of finite type 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 injective. If $\mathcal{F}$ is finitely presented, this canonical morphism is an isomorphism.
Proof.
The map sends the equivalence class of $(U, \varphi )$ in $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})_ x$, where $x \in U \subset X$ is open and $\varphi \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{F}|_ U, \mathcal{G}|_ U)$, to the the induced map on stalks at $x$, namely $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$.
Suppose $\mathcal{F}$ is of finite type. Pick a representative $(U, \varphi )$ of an element $\sigma $ in the kernel of the map, i.e., $\varphi _ x = 0$. Shrinking $U$ if necessary, choose sections $s^1, \ldots , s^ n \in \mathcal{F}(U)$ generating $\mathcal{F}|_ U$. Since $\varphi _ x(s^ i_ x) = 0$ and we are dealing with a finite number of sections, we can find an open neighborhood $V \subset U$ of $x$ such that $\varphi _ V(s^ i|_ V)=0$ for all $i = 1, \ldots , n$. Since $s^ i|_ V$, $i = 1, \ldots , n$ generate $\mathcal{F}|_ V$ this means that $\varphi |_ V = 0$. Since $(U, \varphi )$ is equivalent to $(V, \varphi |_ V)$ we conclude $\sigma = 0$ and injectivity of the map follows.
Next, assume $\mathcal{F}$ is finitely presented. 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 17.22.2 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 17.22.5. 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 17.22.4 and More on Algebra, Lemma 15.65.4 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 17.22.4.
$\square$
Lemma 17.22.6. 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 17.22.4. And the result for coherent sheaves then follows from Lemma 17.12.4.
$\square$
Lemma 17.22.7. Let $X$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation. Let $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } \mathcal{G}_\lambda $ be a filtered colimit of $\mathcal{O}_ X$-modules. Then the canonical map
\[ \mathop{\mathrm{colim}}\nolimits _\lambda \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_\lambda ) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \]
is an isomorphism.
Proof.
Taking colimits of sheaves of modules commutes with restriction to opens, see Sheaves, Section 6.29. Hence we may assume $\mathcal{F}$ has a global presentation
\[ \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{O}_ X \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ X \to \mathcal{F} \to 0 \]
The functor $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(-, -)$ commutes with finite direct sums in either variable and $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, -)$ is the identity functor. By this and by Lemma 17.22.2 we obtain an exact sequence
\[ 0 \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G} \to \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G} \]
Since filtered colimits are exact in $\textit{Mod}(\mathcal{O}_ X)$ also the top row in the following commutative diagram is exact
\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{colim}}\nolimits _\lambda \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}_\lambda ) \ar[r] \ar[d] & \mathop{\mathrm{colim}}\nolimits _\lambda \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G}_\lambda \ar[r] \ar[d] & \mathop{\mathrm{colim}}\nolimits _\lambda \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G}_\lambda \ar[d] \\ 0 \ar[r] & \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) \ar[r] & \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{G} \ar[r] & \bigoplus \nolimits _{j = 1, \ldots , m} \mathcal{G} } \]
Since the right two vertical arrows are isomorphisms we conclude.
$\square$
Lemma 17.22.8. Let $X$ be a ringed space. Let $I$ be a preordered set and let $(\mathcal{F}_ i, \varphi _{ii'})$ be a system over $I$ consisting of sheaves of $\mathcal{O}_ X$-modules (see Categories, Section 4.21). Assume
$I$ is directed,
$\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation, and
$X$ has a cofinal system of open coverings $\mathcal{U} : X = \bigcup _{j\in J} U_ j$ with $J$ finite and $U_ j \cap U_{j'}$ quasi-compact for all $j, j' \in J$.
Then we have
\[ \mathop{\mathrm{colim}}\nolimits _ i \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}_ i) = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i). \]
Proof.
Set $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i)$ and $\mathcal{H}_ i = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{F}_ i)$. Recall that
\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}) = \Gamma (X, \mathcal{H}) \quad \text{and}\quad \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}_ i) = \Gamma (X, \mathcal{H}_ i) \]
by construction. By Lemma 17.22.7 we have $\mathcal{H} = \mathop{\mathrm{colim}}\nolimits \mathcal{H}_ i$. Thus the lemma follows from Sheaves, Lemma 6.29.1.
$\square$
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