Let $X$, $W$ be algebraic spaces with $X$ Noetherian. Let $Z \subset X$ be a closed subspace with open complement $U$. The proposition below says roughly speaking that
\[ \{ \text{morphisms }X \to W\} = \{ \text{compatible morphisms }U \to W\text{ and }X_{/Z} \to W\} \]
where compatibility of $a : U \to W$ and $b : X_{/Z} \to W$ means that $a$ and $b$ define the same “morphism of rig-spaces”. To introduce the category of “rig-spaces” requires a lot of work, but we don't need to do so in order to state precisely what the condition means in this case.
Proposition 88.26.1. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset with complementary open subspace $U \subset X$. Let $f : X' \to X$ be a proper morphism of algebraic spaces such that $f^{-1}(U) \to U$ is an isomorphism. For any algebraic space $W$ over $S$ the map
\[ \mathop{\mathrm{Mor}}\nolimits _ S(X, W) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _ S(X', W) \times _{\mathop{\mathrm{Mor}}\nolimits _ S(X'_{/T}, W)} \mathop{\mathrm{Mor}}\nolimits _ S(X_{/T}, W) \]
is bijective.
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 $n$th 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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