Lemma 96.8.4. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $F$ is algebraic and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces, then $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces.

**Proof.**
Assume $F$ is algebraic and $\Delta : \mathcal{Y} \to \mathcal{Y} \times \mathcal{Y}$ is representable by algebraic spaces. Take a scheme $U$ over $S$ and two objects $x_1, x_2$ of $\mathcal{X}$ over $U$. We have to show that $\mathit{Isom}(x_1, x_2)$ is an algebraic space over $U$, see Algebraic Stacks, Lemma 93.10.11. Set $y_ i = F(x_ i)$. We have a morphism of sheaves of sets

and the target is an algebraic space by assumption. Thus it suffices to show that $f$ is representable by algebraic spaces, see Bootstrap, Lemma 79.3.6. Thus we can choose a scheme $V$ over $U$ and an isomorphism $\beta : y_{1, V} \to y_{2, V}$ and we have to show the functor

is an algebraic space. Consider the objects $z_1 = (V, x_{1, V}, \text{id})$ and $z_2 = (V, x_{2, V}, \beta )$ of

Then it is straightforward to verify that the functor above is equal to $\mathit{Isom}(z_1, z_2)$ on $(\mathit{Sch}/V)_{fppf}$. Hence this is an algebraic space by our assumption that $F$ is algebraic (and the definition of algebraic stacks). $\square$

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