Lemma 102.7.1. Let $g : Z \to Y$ be a morphism of affine schemes. Let $f : \mathcal{X} \to Y$ be a quasi-compact morphism of algebraic stacks. Let $z \in Z$ and let $T \subset |\mathcal{X} \times _ Y Z|$ be a closed subset with $z \not\in \mathop{\mathrm{Im}}(T \to |Z|)$. If $\mathcal{X}$ is quasi-compact, then there exist an open neighbourhood $V \subset Z$ of $z$, 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
$Z'$ is an affine scheme of finite presentation over $Y$,
with $z' = a(z)$ we have $z' \not\in \mathop{\mathrm{Im}}(T' \to |Z'|)$, and
the inverse image of $T$ in $|\mathcal{X} \times _ Y V|$ maps into $T'$ via $|\mathcal{X} \times _ Y V| \to |\mathcal{X} \times _ Y Z'|$.
Proof.
We will deduce this from the corresponding result for morphisms of schemes. Since $\mathcal{X}$ is quasi-compact, we may choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. 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
$Z'$ is an affine scheme of finite presentation over $Y$,
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
\[ \xymatrix{ W \times _ Y Z \ar[d] & W \times _ Y V \ar[l] \ar[rr]_{a_1} \ar[d]_ c & & W \times _ Y Z' \ar[d]^ q \\ \mathcal{X} \times _ Y Z & \mathcal{X} \times _ Y V \ar[l] \ar[rr]^{a_2} & & \mathcal{X} \times _ Y Z' } \]
has cartesian squares and the vertical maps are surjective, smooth, 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 $T_2 = T \cap |\mathcal{X} \times _ Y V|$ by $c$. By Properties of Stacks, Lemma 100.4.3 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 $\overline{a_2(T_2)} \subset |\mathcal{X} \times _ Y Z'|$ as a solution to the problem posed by the lemma.
$\square$
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