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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