The Stacks project

Lemma 78.12.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then we have

  1. The presheaf $(X/Y)_{fin}$ satisfies the sheaf condition for the fppf topology.

  2. If $T$ is an algebraic space over $S$, then there is a canonical bijection

    \[ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})}(T, (X/Y)_{fin}) = \{ (a, Z)\text{ satisfying 04PC}\} \]

Proof. Let $T$ be an algebraic space over $S$. Let $\{ T_ i \to T\} $ be an fppf covering (by algebraic spaces). Let $s_ i = (a_ i, Z_ i)$ be pairs over $T_ i$ satisfying such that we have $s_ i|_{T_ i \times _ T T_ j} = s_ j|_{T_ i \times _ T T_ j}$. First, this implies in particular that $a_ i$ and $a_ j$ define the same morphism $T_ i \times _ T T_ j \to Y$. By Descent on Spaces, Lemma 73.7.2 we deduce that there exists a unique morphism $a : T \to Y$ such that $a_ i$ equals the composition $T_ i \to T \to Y$. Second, this implies that $Z_ i \subset T_ i \times _ Y X$ are open subspaces whose inverse images in $(T_ i \times _ T T_ j) \times _ Y X$ are equal. Since $\{ T_ i \times _ Y X \to T \times _ Y X\} $ is an fppf covering we deduce that there exists a unique open subspace $Z \subset T \times _ Y X$ which restricts back to $Z_ i$ over $T_ i$, see Descent on Spaces, Lemma 73.7.1. We claim that the projection $Z \to T$ is finite. This follows as being finite is local for the fpqc topology, see Descent on Spaces, Lemma 73.11.23.

Note that the result of the preceding paragraph in particular implies (1).

Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say “pair” we mean a pair satisfying conditions

Let $v : T \to (X/Y)_{fin}$ be a natural transformation. Choose a scheme $U$ and a surjective étale morphism $p : U \to T$. Then $v(p) \in (X/Y)_{fin}(U)$ corresponds to a pair $(a_ U, Z_ U)$ over $U$. Let $R = U \times _ T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_ U, Z_ U)$ by $s$ and $t$ agree. Hence, since $\{ U \to T\} $ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, Z)$ over $T$.

Conversely, let $(a, Z)$ be a pair over $T$. Let $U \to T$, $R = U \times _ T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, Z)|_ U$ gives rise to a transformation of functors $v : h_ U \to (X/Y)_{fin}$ by the Yoneda lemma (Categories, Lemma 4.3.5). As the two pullbacks $s^*(a, Z)|_ U$ and $t^*(a, Z)|_ U$ are equal, we see that $v$ coequalizes the two maps $h_ t, h_ s : h_ R \to h_ U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma 64.9.1 and since $(X/Y)_{fin}$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to (X/Y)_{fin}$.

We omit the verification that the two constructions above are mutually inverse. $\square$

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