Lemma 65.27.3. Let $S$ be a scheme and let $Y$ be an algebraic space over $S$. Let $\mathcal{F}$ be a sheaf of sets on $Y_{\acute{e}tale}$. Provided a set theoretic condition is satisfied (see proof) the functor $X$ associated to $\mathcal{F}$ above is an algebraic space and there is an étale morphism $f : X \to Y$ of algebraic spaces such that $\mathcal{F} = f_{small, *}*$ where $*$ is the final object of the category $\mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$ (constant sheaf with value a singleton).

Proof. Let us prove that $X$ is a sheaf for the fppf topology. Namely, suppose that $\{ g_ i : T_ i \to T\}$ is a covering of $(\mathit{Sch}/S)_{fppf}$ and $(y_ i, s_ i) \in X(T_ i)$ satisfy the glueing condition, i.e., the restriction of $(y_ i, s_ i)$ and $(y_ j, s_ j)$ to $T_ i \times _ T T_ j$ agree. Then since $Y$ is a sheaf for the fppf topology, we see that the $y_ i$ give rise to a unique morphism $y : T \to Y$ such that $y_ i = y \circ g_ i$. Then we see that $y_{i, small}^{-1}\mathcal{F} = g_{i, small}^{-1}y_{small}^{-1}\mathcal{F}$. Hence the sections $s_ i$ glue uniquely to a section of $y_{small}^{-1}\mathcal{F}$ by Étale Cohomology, Lemma 59.39.2.

The construction that sends $\mathcal{F} \in \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}))$ to $X \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ preserves finite limits and all colimits since each of the functors $y_{small}^{-1}$ have this property. Of course, if $V \in \mathop{\mathrm{Ob}}\nolimits (Y_{\acute{e}tale})$, then the construction sends the representable sheaf $h_ V$ on $Y_{\acute{e}tale}$ to the representable functor represented by $V$.

By Sites, Lemma 7.12.5 we can find a set $I$, for each $i \in I$ an object $V_ i$ of $Y_{\acute{e}tale}$ and a surjective map of sheaves

$\coprod h_{V_ i} \longrightarrow \mathcal{F}$

on $Y_{\acute{e}tale}$. The set theoretic condition we need is that the index set $I$ is not too large1. Then $V = \coprod V_ i$ is an object of $(\mathit{Sch}/S)_{fppf}$ and therefore an object of $Y_{\acute{e}tale}$ and we have a surjective map $h_ V \to \mathcal{F}$.

Observe that the product of $h_ V$ with itself in $\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale})$ is $h_{V \times _ Y V}$. Consider the fibre product

$h_ V \times _\mathcal {F} h_ V \subset h_{V \times _ Y V}$

There is an open subscheme $R$ of $V \times _ Y V$ such that $h_ V \times _\mathcal {F} h_ V = h_ R$, see Lemma 65.20.1 (small detail omitted). By the Yoneda lemma we obtain two morphisms $s, t : R \to V$ in $Y_{\acute{e}tale}$ and we find a coequalizer diagram

$\xymatrix{ h_ R \ar@<1ex>[r] \ar@<-1ex>[r] & h_ V \ar[r] & \mathcal{F} }$

in $\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale})$. Of course the morphisms $s, t$ are étale and define an étale equivalence relation $(t, s) : R \to V \times _ S V$.

By the discussion in the preceding two paragraphs we find a coequalizer diagram

$\xymatrix{ R \ar@<1ex>[r] \ar@<-1ex>[r] & V \ar[r] & X }$

in $(\mathit{Sch}/S)_{fppf}$. Thus $X = V/R$ is an algebraic space by Spaces, Theorem 64.10.5. The other statements follow readily from this; details omitted. $\square$

[1] It suffices if the supremum of the cardinalities of the stalks of $\mathcal{F}$ at geometric points of $Y$ is bounded by the size of some object of $(\mathit{Sch}/S)_{fppf}$.

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