The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01CM. Beware of the difference between the letter 'O' and the digit '0'.