Lemma 99.3.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $i : X' \to X$ be a closed immersion of algebraic spaces over $B$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module and let $\mathcal{G}'$ be a quasi-coherent $\mathcal{O}_{X'}$-module. Then

$\mathit{Hom}(\mathcal{F}, i_*\mathcal{G}') = \mathit{Hom}(i^*\mathcal{F}, \mathcal{G}')$

as functors on $(\mathit{Sch}/B)$.

Proof. Let $g : T \to B$ be a morphism where $T$ is a scheme. Denote $i_ T : X'_ T \to X_ T$ the base change of $i$. Denote $h : X_ T \to X$ and $h' : X'_ T \to X'$ the projections. Observe that $(h')^*i^*\mathcal{F} = i_ T^*h^*\mathcal{F}$. As a closed immersion is affine (Morphisms of Spaces, Lemma 67.20.6) we have $h^*i_*\mathcal{G} = i_{T, *}(h')^*\mathcal{G}$ by Cohomology of Spaces, Lemma 69.11.1. Thus we have

\begin{align*} \mathit{Hom}(\mathcal{F}, i_*\mathcal{G}')(T) & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_ T}}(h^*\mathcal{F}, h^*i_*\mathcal{G}') \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_ T}}(h^*\mathcal{F}, i_{T, *}(h')^*\mathcal{G}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'_ T}}(i_ T^*h^*\mathcal{F}, (h')^*\mathcal{G}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'_ T}}((h')^*i^*\mathcal{F}, (h')^*\mathcal{G}) \\ & = \mathit{Hom}(i^*\mathcal{F}, \mathcal{G}')(T) \end{align*}

as desired. The middle equality follows from the adjointness of the functors $i_{T, *}$ and $i_ T^*$. $\square$

## Comments (0)

There are also:

• 2 comment(s) on Section 99.3: The Hom functor

## 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Your email address will not be published.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 08K5. Beware of the difference between the letter 'O' and the digit '0'.