In Situation 81.10.6 we consider the category $\textit{Spaces}(Y \to X, Z)$ of commutative diagrams of algebraic spaces over $S$ of the form

\[ \xymatrix{ U' \ar[d] & V' \ar[l] \ar[d] \ar[r] & Y' \ar[d] \\ U & V \ar[l] \ar[r] & Y } \]

where both squares are cartesian. There is a canonical functor

81.11.0.1
\begin{equation} \label{spaces-pushouts-equation-formal-glueing-spaces} \textit{Spaces}/X \longrightarrow \textit{Spaces}(Y \to X, Z) \end{equation}

which maps $X' \to X$ to the morphisms $U \times _ X X' \leftarrow V \times _ X X' \rightarrow Y \times _ X X'$.

Lemma 81.11.1. In Situation 81.10.6 the functor (81.11.0.1) restricts to an equivalence

from the category of algebraic spaces affine over $X$ to the full subcategory of $\textit{Spaces}(Y \to X, Z)$ consisting of $(U' \leftarrow V' \rightarrow Y')$ with $U' \to U$, $V' \to V$, and $Y' \to Y$ affine,

from the category of closed immersions $X' \to X$ to the full subcategory of $\textit{Spaces}(Y \to X, Z)$ consisting of $(U' \leftarrow V' \rightarrow Y')$ with $U' \to U$, $V' \to V$, and $Y' \to Y$ closed immersions, and

same statement as in (2) for finite morphisms.

**Proof.**
The category of algebraic spaces affine over $X$ is equivalent to the category of quasi-coherent sheaves $\mathcal{A}$ of $\mathcal{O}_ X$-algebras. The full subcategory of $\textit{Spaces}(Y \to X, Z)$ consisting of $(U' \leftarrow V' \rightarrow Y')$ with $U' \to U$, $V' \to V$, and $Y' \to Y$ affine is equivalent to the category of algebra objects of $\mathit{QCoh}(Y \to X, Z)$. In both cases this follows from Morphisms of Spaces, Lemma 67.20.7 with quasi-inverse given by the relative spectrum construction (Morphisms of Spaces, Definition 67.20.8) which commutes with arbitrary base change. Thus part (1) of the lemma follows from Proposition 81.10.9.

Fully faithfulness in part (2) follows from part (1). For essential surjectivity, we reduce by part (1) to proving that $X' \to X$ is a closed immersion if and only if both $U \times _ X X' \to U$ and $Y \times _ X X' \to Y$ are closed immersions. By Lemma 81.10.11 $\{ U \to X, Y \to X\} $ can be refined by an fpqc covering. Hence the result follows from Descent on Spaces, Lemma 74.11.17.

For (3) use the argument proving (2) and Descent on Spaces, Lemma 74.11.23.
$\square$

Lemma 81.11.2. In Situation 81.10.6 the functor (81.11.0.1) reflects isomorphisms.

**Proof.**
By a formal argument with base change, this reduces to the following question: A morphism $a : X' \to X$ of algebraic spaces such that $U \times _ X X' \to U$ and $Y \times _ X X' \to Y$ are isomorphisms, is an isomorphism. The family $\{ U \to X, Y \to X\} $ can be refined by an fpqc covering by Lemma 81.10.11. Hence the result follows from Descent on Spaces, Lemma 74.11.15.
$\square$

Lemma 81.11.3. In Situation 81.10.6 the functor (81.11.0.1) is fully faithful on algebraic spaces separated over $X$. More precisely, it induces a bijection

\[ \mathop{\mathrm{Mor}}\nolimits _ X(X'_1, X'_2) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\textit{Spaces}(Y \to X, Z)}(F(X'_1), F(X'_2)) \]

whenever $X'_2 \to X$ is separated.

**Proof.**
Since $X'_2 \to X$ is separated, the graph $i : X'_1 \to X'_1 \times _ X X'_2$ of a morphism $X'_1 \to X'_2$ over $X$ is a closed immersion, see Morphisms of Spaces, Lemma 67.4.6. Moreover a closed immersion $i : T \to X'_1 \times _ X X'_2$ is the graph of a morphism if and only if $\text{pr}_1 \circ i$ is an isomorphism. The same is true for

the graph of a morphism $U \times _ X X'_1 \to U \times _ X X'_2$ over $U$,

the graph of a morphism $V \times _ X X'_1 \to V \times _ X X'_2$ over $V$, and

the graph of a morphism $Y \times _ X X'_1 \to Y \times _ X X'_2$ over $Y$.

Moreover, if morphisms as in (1), (2), (3) fit together to form a morphism in the category $\textit{Spaces}(Y \to X, Z)$, then these graphs fit together to give an object of $\textit{Spaces}(Y \times _ X (X'_1 \times _ X X'_2) \to X'_1 \times _ X X'_2, Z \times _ X (X'_1 \times _ X X'_2))$ whose triple of morphisms are closed immersions. The proof is finished by applying Lemmas 81.11.1 and 81.11.2.
$\square$

## Comments (2)

Comment #7780 by Laurent Moret-Bailly on

Comment #8021 by Stacks Project on