Proof.
Let w' : X' \to W and \hat w : X_{/T} \to W be morphisms which determine the same morphism X'_{/T} \to W by composition with X'_{/T} \to X and X'_{/T} \to X_{/T}. We have to prove there exists a unique morphism w : X \to W whose composition with X' \to X and X_{/T} \to X recovers w' and \hat w. The uniqueness is immediate from Lemma 88.25.4.
The assumptions on T and f are preserved by base change by any étale morphism X_1 \to X of algebraic spaces. Since formal algebraic spaces are sheaves for the étale topology and since we already have the uniqueness, it suffices to prove existence after replacing X by the members of an étale covering. Thus we may assume X is an affine Noetherian scheme.
Assume X is an affine Noetherian scheme. We will construct the morphism w : X \to W using the material in Pushouts of Spaces, Section 81.13. It makes sense to read a little bit of the material in that section before continuing the read the proof.
Set X'' = X' \times _ X X' and consider the two morphisms a = w' \circ \text{pr}_1 : X'' \to W and b = w' \circ \text{pr}_2 : X'' \to W. Then we see that a and b agree over the open U and that a_{/T} and b_{a/T} agree (as these are both equal to the composition X''_{/T} \to X_{/T} \to W where the second arrow is \hat w). Thus by Lemma 88.25.4 we see a = b.
Denote Z \subset X the reduced induced closed subscheme structure on T. For n \geq 1 denote Z_ n \subset X the nth infinitesimal neighbourhood of Z. Denote w_ n = \hat w|_{Z_ n} : Z_ n \to W so that we have \hat w = \mathop{\mathrm{colim}}\nolimits w_ n on X_{/T} = \mathop{\mathrm{colim}}\nolimits Z_ n. Set Y_ n = X' \amalg Z_ n. Consider the two projections
s_ n, t_ n : R_ n = Y_ n \times _ X Y_ n \longrightarrow Y_ n
Let Y_ n \to X_ n \to X be the coequalizer of s_ n and t_ n as in Pushouts of Spaces, Section 81.13 (in particular this coequalizer exists, has good properties, etc, see Pushouts of Spaces, Lemma 81.13.1). By the result a = b of the previous parapgraph and the agreement of w' and \hat w over X'_{/T} we see that the morphism
w' \amalg w_ n : Y_ n \longrightarrow W
equalizes the morphisms s_ n and t_ n. Hence we see that for all n \geq 1 there is a morphism w^ n : X_ n \to W compatible with w' and w_ n. Moreover, for m \geq 1 the composition
X_ n \to X_{n + m} \xrightarrow {w^{n + m}} W
is equal to w^ n by construction (as the corresponding statement holds for w' \amalg w_{n + m} and w' \amalg w_ n). By Pushouts of Spaces, Lemma 81.13.4 and Remark 81.13.5 the system of algebraic spaces X_ n is essentially constant with value X and we conclude.
\square
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