The Stacks project

Proof. By Lemma 99.3.2 the functor $\mathit{Hom}(\mathcal{F}, \mathcal{G})$ satisfies the sheaf property for fppf coverings. This mean we may¹ apply Bootstrap, Lemma 80.11.1 to check the representability étale locally on $B$. Moreover, we may check whether the end result is affine or of finite presentation étale locally on $B$, see Morphisms of Spaces, Lemmas 67.20.3 and 67.28.4. Hence we may assume that $B$ is an affine scheme.

Assume $B$ is an affine scheme. As $f$ is of finite presentation, it follows $X$ is quasi-compact and quasi-separated. Thus we can write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i$ as a filtered colimit of $\mathcal{O}_ X$-modules of finite presentation (Limits of Spaces, Lemma 70.9.1). It is clear that

Hence if we can show that each $\mathit{Hom}(\mathcal{F}_ i, \mathcal{G})$ is representable by an affine scheme, then we see that the same thing holds for $\mathit{Hom}(\mathcal{F}, \mathcal{G})$. Use the material in Limits, Section 32.2 and Limits of Spaces, Section 70.4. Thus we may assume that $\mathcal{F}$ is of finite presentation.

Say $B = \mathop{\mathrm{Spec}}(R)$. Write $R = \mathop{\mathrm{colim}}\nolimits R_ i$ with each $R_ i$ a finite type $\mathbf{Z}$-algebra. Set $B_ i = \mathop{\mathrm{Spec}}(R_ i)$. By the results of Limits of Spaces, Lemmas 70.7.1 and 70.7.2 we can find an $i$, a morphism of algebraic spaces $X_ i \to B_ i$, and finitely presented $\mathcal{O}_{X_ i}$-modules $\mathcal{F}_ i$ and $\mathcal{G}_ i$ such that the base change of $(X_ i, \mathcal{F}_ i, \mathcal{G}_ i)$ to $B$ recovers $(X, \mathcal{F}, \mathcal{G})$. By Limits of Spaces, Lemma 70.6.12 we may, after increasing $i$, assume that $\mathcal{G}_ i$ is flat over $B_ i$. By Limits of Spaces, Lemma 70.12.3 we may similarly assume the scheme theoretic support of $\mathcal{G}_ i$ is proper over $B_ i$. At this point we can apply Lemma 99.3.9 to see that $H_ i = \mathit{Hom}(\mathcal{F}_ i, \mathcal{G}_ i)$ is an algebraic space affine of finite presentation over $B_ i$. Pulling back to $B$ (using Remark 99.3.4) we see that $H_ i \times _{B_ i} B = \mathit{Hom}(\mathcal{F}, \mathcal{G})$ and we win. $\square$