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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.

  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. 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

  1. $I$ is directed,

  2. $\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation, and

  3. $X$ 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. $\square$

Remark 17.22.9. In the lemma above some condition beyond the condition that $X$ is quasi-compact is necessary. See Sheaves, Example 6.29.2.


Comments (1)

Comment #9483 by Huang on

Should Lemma 17.22.2.(1) be correct only for being coherent(or finite presentation), then it follows from the stalk isomorphism of intern Hom, while the general case the proof would not go through since surjective of sheaves only implies surjective on the stalk level?


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