Lemma 35.13.8. Consider schemes $X, Y, Z$ and morphisms $a, b : X \to Y$ and a morphism $c : Y \to Z$ with $c \circ a = c \circ b$. Set $d = c \circ a = c \circ b$. If there exists an fpqc covering $\{ Z_ i \to Z\}$ such that

1. for all $i$ the morphism $Y \times _{c, Z} Z_ i \to Z_ i$ is the coequalizer of $(a, 1) : X \times _{d, Z} Z_ i \to Y \times _{c, Z} Z_ i$ and $(b, 1) : X \times _{d, Z} Z_ i \to Y \times _{c, Z} Z_ i$, and

2. for all $i$ and $i'$ the morphism $Y \times _{c, Z} (Z_ i \times _ Z Z_{i'}) \to (Z_ i \times _ Z Z_{i'})$ is the coequalizer of $(a, 1) : X \times _{d, Z} (Z_ i \times _ Z Z_{i'}) \to Y \times _{c, Z} (Z_ i \times _ Z Z_{i'})$ and $(b, 1) : X \times _{d, Z} (Z_ i \times _ Z Z_{i'}) \to Y \times _{c, Z} (Z_ i \times _ Z Z_{i'})$

then $c$ is the coequalizer of $a$ and $b$.

Proof. Namely, for a scheme $T$ a morphism $Z \to T$ is the same thing as a collection of morphism $Z_ i \to T$ which agree on overlaps by Lemma 35.13.7. $\square$

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