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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)