Lemma 37.64.6. In Situation 37.64.1 suppose given a commutative diagram

\[ \xymatrix{ Y' \ar[d]^ g & Z' \ar[l]^{j'} \ar[r]_{i'} \ar[d]^ h & X' \ar[d]^ f \\ Y & Z \ar[l] \ar[r] & X } \]

with cartesian squares and $f, g, h$ separated and locally quasi-finite. Then

the pushouts $Y \amalg _ Z X$ and $Y' \amalg _{Z'} X'$ exist,

$Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is separated and locally quasi-finite, and

the squares

\[ \xymatrix{ Y' \ar[r] \ar[d] & Y' \amalg _{Z'} X' \ar[d] & X' \ar[l] \ar[d] \\ Y \ar[r] & Y \amalg _ Z X & X \ar[l] } \]

are cartesian.

**Proof.**
The pushout $Y \amalg _ Z X$ exists by Proposition 37.64.3. To see that the pushout $Y' \amalg _{Z'} X'$ exists, we check condition (3) of Situation 37.64.1 holds for $(X', Y', Z', i', j')$. Namely, let $y' \in Y'$ and denote $y \in Y$ the image. Choose $U \subset X$ affine open with $i(j^{-1}(y)) \subset U$. Choose a quasi-compact open $U' \subset X'$ contained in $f^{-1}(U)$ containing the quasi-compact subset $i'((j')^{-1}(\{ y'\} ))$. By Lemma 37.63.8 we see that $U'$ is quasi-affine. Since $Z'_{y'}$ is the spectrum of an algebra integral over a field, we can apply Limits, Lemma 32.11.6 and we find there exists an affine open subscheme of $U'$ containing $i'((j')^{-1}(\{ y'\} ))$ as desired.

Having verified existence we check the other assertions. Affine locally we are exactly in the situation of More on Algebra, Lemma 15.7.7 with $B \to D$ and $A' \to C'$ locally quasi-finite^{1}. In particular, the morphism $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is locally of finite type. The squares in of the diagram are cartesian by More on Algebra, Lemma 15.6.4. Since being locally quasi-finite can be checked on fibres (Morphisms, Lemma 29.20.6) we conclude that $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is locally quasi-finite.

We still have to check $Y' \amalg _{Z'} X' \to Y \amalg _ Z X$ is separated. Observe that $Y' \amalg X' \to Y' \amalg _{Z'} X'$ is universally closed and surjective by Proposition 37.64.3. Since also the morphism $Y' \amalg X' \to Y \amalg _ Z X$ is separated (as it factors as $Y' \amalg X' \to Y \amalg X \to Y \amalg _ Z X$) we conclude by Morphisms, Lemma 29.41.11.
$\square$

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