Lemma 69.20.1. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Z \to Y$ be morphisms of algebraic spaces over $S$. Let $z \in |Z|$ and let $T \subset |X \times _ Y Z|$ be a closed subset with $z \not\in \mathop{\mathrm{Im}}(T \to |Z|)$. If $f$ is quasi-compact, then there exists an étale neighbourhood $(V, v) \to (Z, z)$, a commutative diagram

\[ \xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, } \]

and a closed subset $T' \subset |X \times _ Y Z'|$ such that

the morphism $b : Z' \to Y$ is locally of finite presentation,

with $z' = a(v)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and

the inverse image of $T$ in $|X \times _ Y V|$ maps into $T'$ via $|X \times _ Y V| \to |X \times _ Y Z'|$.

Moreover, we may assume $V$ and $Z'$ are affine schemes and if $Z$ is a scheme we may assume $V$ is an affine open neighbourhood of $z$.

**Proof.**
We will deduce this from the corresponding result for morphisms of schemes. Let $y \in |Y|$ be the image of $z$. First we choose an affine étale neighbourhood $(U, u) \to (Y, y)$ and then we choose an affine étale neighbourhood $(V, v) \to (Z, z)$ such that the morphism $V \to Y$ factors through $U$. Then we may replace

$X \to Y$ by $X \times _ Y U \to U$,

$Z \to Y$ by $V \to U$,

$z$ by $v$, and

$T$ by its inverse image in $|(X \times _ Y U) \times _ U V| = |X \times _ Y V|$.

In fact, below we will show that after replacing $V$ by an affine open neighbourhood of $v$ there will be a morphism $a : V \to Z'$ for some $Z' \to U$ of finite presentation and a closed subset $T'$ of $|(X \times _ Y U) \times _ U Z'| = |X \times _ Y Z'|$ such that $T$ maps into $T'$ and $a(v) \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$. Thus we may and do assume that $Z$ and $Y$ are affine schemes with the proviso that we need to find a solution where $V$ is an open neighbourhood of $z$.

Since $f$ is quasi-compact and $Y$ is affine, the algebraic space $X$ is quasi-compact. Choose an affine scheme $W$ and a surjective étale morphism $W \to X$. Let $T_ W \subset |W \times _ Y Z|$ be the inverse image of $T$. Then $z$ is not in the image of $T_ W$. By the schemes case (Limits, Lemma 32.14.1) we can find an open neighbourhood $V \subset Z$ of $z$ a commutative diagram of schemes

\[ \xymatrix{ V \ar[d] \ar[r]_ a & Z' \ar[d]^ b \\ Z \ar[r]^ g & Y, } \]

and a closed subset $T' \subset |W \times _ Y Z'|$ such that

the morphism $b : Z' \to Y$ is locally of finite presentation,

with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to Z')$, and

$T_1 = T_ W \cap |W \times _ Y V|$ maps into $T'$ via $|W \times _ Y V| \to |W \times _ Y Z'|$.

The commutative diagram

\[ \xymatrix{ W \times _ Y Z \ar[d] & W \times _ Y V \ar[l] \ar[rr]_{a_1} \ar[d]_ c & & W \times _ Y Z' \ar[d]^ q \\ X \times _ Y Z & X \times _ Y V \ar[l] \ar[rr]^{a_2} & & X \times _ Y Z' } \]

has cartesian squares and the vertical maps are, surjective, étale and a fortiori open. Looking at the left hand square we see that $T_1 = T_ W \cap |W \times _ Y V|$ is the inverse image of $T_2 = T \cap |X \times _ Y V|$ by $c$. By Properties of Spaces, Lemma 65.4.3 we get $a_1(T_1) = q^{-1}(a_2(T_2))$. By Topology, Lemma 5.6.4 we get

\[ q^{-1}\left(\overline{a_2(T_2)}\right) = \overline{q^{-1}(a_2(T_2))} = \overline{a_1(T_1)} \subset T' \]

As $q$ is surjective the image of $\overline{a_2(T_2)} \to |Z'|$ does not contain $z'$ since the same is true for $T'$. Thus we can take the diagram with $Z', V, a, b$ above and the closed subset $\overline{a_2(T_2)} \subset |X \times _ Y Z'|$ as a solution to the problem posed by the lemma.
$\square$

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