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
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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