Lemma 100.28.2. Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}$ is a gerbe, then the sheafification of the presheaf

$(\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}, \quad U \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)/\! \! \cong$

is an algebraic space and $\mathcal{X}$ is a gerbe over it.

Proof. (In this proof the abuse of language introduced in Section 100.2 really pays off.) Choose a morphism $\pi : \mathcal{X} \to X$ where $X$ is an algebraic space which turns $\mathcal{X}$ into a gerbe over $X$. It suffices to prove that $X$ is the sheafification of the presheaf $\mathcal{F}$ displayed in the lemma. It is clear that there is a map $c : \mathcal{F} \to X$. We will use Stacks, Lemma 8.11.3 properties (2)(a) and (2)(b) to see that the map $c^\# : \mathcal{F}^\# \to X$ is surjective and injective, hence an isomorphism, see Sites, Lemma 7.11.2. Surjective: Let $T$ be a scheme and let $f : T \to X$. By property (2)(a) there exists an fppf covering $\{ h_ i : T_ i \to T\}$ and morphisms $x_ i : T_ i \to \mathcal{X}$ such that $f \circ h_ i$ corresponds to $\pi \circ x_ i$. Hence we see that $f|_{T_ i}$ is in the image of $c$. Injective: Let $T$ be a scheme and let $x, y : T \to \mathcal{X}$ be morphisms such that $c \circ x = c \circ y$. By (2)(b) we can find a covering $\{ T_ i \to T\}$ and morphisms $x|_{T_ i} \to y|_{T_ i}$ in the fibre category $\mathcal{X}_{T_ i}$. Hence the restrictions $x|_{T_ i}, y|_{T_ i}$ are equal in $\mathcal{F}(T_ i)$. This proves that $x, y$ give the same section of $\mathcal{F}^\#$ over $T$ as desired. $\square$

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