Lemma 99.5.3. In Situation 99.5.1. Denote $\mathcal{X} = \mathcal{C}\! \mathit{oh}_{X/B}$. Then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces.
Proof. Consider two objects $x = (T, g, \mathcal{F})$ and $y = (T, h, \mathcal{G})$ of $\mathcal{X}$ over a scheme $T$. We have to show that $\mathit{Isom}_\mathcal {X}(x, y)$ is an algebraic space over $T$, see Algebraic Stacks, Lemma 94.10.11. If for $a : T' \to T$ the restrictions $x|_{T'}$ and $y|_{T'}$ are isomorphic in the fibre category $\mathcal{X}_{T'}$, then $g \circ a = h \circ a$. Hence there is a transformation of presheaves
\[ \mathit{Isom}_\mathcal {X}(x, y) \longrightarrow \text{Equalizer}(g, h) \]
Since the diagonal of $B$ is representable (by schemes) this equalizer is a scheme. Thus we may replace $T$ by this equalizer and the sheaves $\mathcal{F}$ and $\mathcal{G}$ by their pullbacks. Thus we may assume $g = h$. In this case we have $\mathit{Isom}_\mathcal {X}(x, y) = \mathit{Isom}(\mathcal{F}, \mathcal{G})$ and the result follows from Proposition 99.4.3. $\square$
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