Lemma 66.14.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $Y, Z \subset X$ be closed subspaces. Let $Y \cup Z$ be the scheme theoretic union of $Y$ and $Z$. Let $Y \cap Z$ be the scheme theoretic intersection of $Y$ and $Z$. Then $Y \to Y \cup Z$ and $Z \to Y \cup Z$ are closed immersions, there is a short exact sequence

\[ 0 \to \mathcal{O}_{Y \cup Z} \to \mathcal{O}_ Y \times \mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z} \to 0 \]

of $\mathcal{O}_ Z$-modules, and the diagram

\[ \xymatrix{ Y \cap Z \ar[r] \ar[d] & Y \ar[d] \\ Z \ar[r] & Y \cup Z } \]

is cocartesian in the category of algebraic spaces over $S$, i.e., $Y \cup Z = Y \amalg _{Y \cap Z} Z$.

**Proof.**
The morphisms $Y \to Y \cup Z$ and $Z \to Y \cup Z$ are closed immersions by Lemma 66.13.1. In the short exact sequence we use the equivalence of Lemma 66.14.1 to think of quasi-coherent modules on closed subspaces of $X$ as quasi-coherent modules on $X$. For the first map in the sequence we use the canonical maps $\mathcal{O}_{Y \cup Z} \to \mathcal{O}_ Y$ and $\mathcal{O}_{Y \cup Z} \to \mathcal{O}_ Z$ and for the second map we use the canonical map $\mathcal{O}_ Y \to \mathcal{O}_{Y \cap Z}$ and the negative of the canonical map $\mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z}$. Then to check exactness we may work étale locally and deduce exactness from the case of schemes (Morphisms, Lemma 29.4.6).

To show the diagram is cocartesian, suppose we are given an algebraic space $T$ over $S$ and morphisms $f : Y \to T$, $g : Z \to T$ agreeing as morphisms $Y \cap Z \to T$. Goal: Show there exists a unique morphism $h : Y \cup Z \to T$ agreeing with $f$ and $g$. To construct $h$ we may work étale locally on $Y \cup Z$ (as $Y \cup Z$ is an étale sheaf being an algebraic space). Hence we may assume that $X$ is a scheme. In this case we know that $Y \cup Z$ is the pushout of $Y$ and $Z$ along $Y \cap Z$ in the category of schemes by Morphisms, Lemma 29.4.6. Choose a scheme $T'$ and a surjective étale morphism $T' \to T$. Set $Y' = T' \times _{T, f} Y$ and $Z' = T' \times _{T, g} Z$. Then $Y'$ and $Z'$ are schemes and we have a canonical isomorphism $\varphi : Y' \times _ Y (Y \cap Z) \to Z' \times _ Z (Y \cap Z)$ of schemes. By More on Morphisms, Lemma 37.65.8 the pushout $W' = Y' \amalg _{Y' \times _ Y (Y \cap Z), \varphi } Z'$ exists in the category of schemes. The morphism $W' \to Y \cup Z$ is étale by More on Morphisms, Lemma 37.65.9. It is surjective as $Y' \to Y$ and $Z' \to Z$ are surjective. The morphisms $f' : Y' \to T'$ and $g' : Z' \to T'$ glue to a unique morphism of schemes $h' : W' \to T'$. By uniqueness the composition $W' \to T' \to T$ descends to the desired morphism $h : Y \cup Z \to T$. Some details omitted.
$\square$

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