Lemma 70.20.1. 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 66.4.3 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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