## 99.4 The Isom functor

In Situation 99.3.1 we can consider the subfunctor

$\mathit{Isom}(\mathcal{F}, \mathcal{G}) \subset \mathit{Hom}(\mathcal{F}, \mathcal{G})$

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.

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.

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

$\mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})}(T, \mathit{Isom}(\mathcal{F}, \mathcal{G})) = \{ (h, u) \mid h : T \to B, u : \mathcal{F}_ T \to \mathcal{G}_ T\text{ isomorphism}\}$

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

1. $f$ is of finite presentation, and

2. $\mathcal{F}$ 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

$c : H' \times _ B H \longrightarrow I \times _ B I',\quad (u', u) \longmapsto (u \circ u', u' \circ u)$

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

$\mathit{Isom}(\mathcal{F}, \mathcal{G}) = (H' \times _ B H) \times _{c, I \times _ B I', \sigma } B$

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

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).