Lemma 99.4.1. In Situation 99.3.1 the functor $\mathit{Isom}(\mathcal{F}, \mathcal{G})$ satisfies the sheaf property for the fpqc topology.

## 99.4 The Isom functor

In Situation 99.3.1 we can consider the subfunctor

whose value on a scheme $T$ over $B$ is the set of *invertible* $\mathcal{O}_{X_ T}$-homomorphisms $u : \mathcal{F}_ T \to \mathcal{G}_ T$.

We sometimes think of $\mathit{Isom}(\mathcal{F}, \mathcal{G})$ as a functor $(\mathit{Sch}/S)^{opp} \to \textit{Sets}$ endowed with a morphism $\mathit{Isom}(\mathcal{F}, \mathcal{G}) \to B$. Namely, if $T$ is a scheme over $S$, then an element of $\mathit{Isom}(\mathcal{F}, \mathcal{G})(T)$ consists of a pair $(h, u)$, where $h$ is a morphism $h : T \to B$ and $u : \mathcal{F}_ T \to \mathcal{G}_ T$ is an $\mathcal{O}_{X_ T}$-module isomorphism where $X_ T = T \times _{h, B} X$ and $\mathcal{F}_ T$ and $\mathcal{G}_ T$ are the pullbacks to $X_ T$. In particular, when we say that $\mathit{Isom}(\mathcal{F}, \mathcal{G})$ is an algebraic space, we mean that the corresponding functor $(\mathit{Sch}/S)^{opp} \to \textit{Sets}$ is an algebraic space.

**Proof.**
We have already seen that $\mathit{Hom}(\mathcal{F}, \mathcal{G})$ satisfies the sheaf property. Hence it remains to show the following: Given an fpqc covering $\{ T_ i \to T\} _{i \in I}$ of schemes over $B$ and an $\mathcal{O}_{X_ T}$-linear map $u : \mathcal{F}_ T \to \mathcal{G}_ T$ such that $u_{T_ i}$ is an isomorphism for all $i$, then $u$ is an isomorphism. Since $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$, see Topologies on Spaces, Lemma 73.9.3, this follows from Descent on Spaces, Proposition 74.4.1.
$\square$

Sanity check: $\mathit{Isom}$ sheaf plays the same role among algebraic spaces over $S$.

Lemma 99.4.2. In Situation 99.3.1. Let $T$ be an algebraic space over $S$. We have

where $\mathcal{F}_ T, \mathcal{G}_ T$ denote the pullbacks of $\mathcal{F}$ and $\mathcal{G}$ to the algebraic space $X \times _{B, h} T$.

**Proof.**
Observe that the left and right hand side of the equality are subsets of the left and right hand side of the equality in Lemma 99.3.3. We omit the verification that these subsets correspond under the identification given in the proof of that lemma.
$\square$

Proposition 99.4.3. In Situation 99.3.1 assume that

$f$ is of finite presentation, and

$\mathcal{F}$ and $\mathcal{G}$ are finitely presented $\mathcal{O}_ X$-modules, flat over $B$, with support proper over $B$.

Then the functor $\mathit{Isom}(\mathcal{F}, \mathcal{G})$ is an algebraic space affine of finite presentation over $B$.

**Proof.**
We will use the abbreviations $H = \mathit{Hom}(\mathcal{F}, \mathcal{G})$, $I = \mathit{Hom}(\mathcal{F}, \mathcal{F})$, $H' = \mathit{Hom}(\mathcal{G}, \mathcal{F})$, and $I' = \mathit{Hom}(\mathcal{G}, \mathcal{G})$. By Proposition 99.3.10 the functors $H$, $I$, $H'$, $I'$ are algebraic spaces and the morphisms $H \to B$, $I \to B$, $H' \to B$, and $I' \to B$ are affine and of finite presentation. The composition of maps gives a morphism

of algebraic spaces over $B$. Since $I \times _ B I' \to B$ is separated, the section $\sigma : B \to I \times _ B I'$ corresponding to $(\text{id}_\mathcal {F}, \text{id}_\mathcal {G})$ is a closed immersion (Morphisms of Spaces, Lemma 67.4.7). Moreover, $\sigma $ is of finite presentation (Morphisms of Spaces, Lemma 67.28.9). Hence

is an algebraic space affine of finite presentation over $B$ as well. Some details omitted. $\square$

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