Lemma 76.18.1. Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \subset Y')$ be a morphism of first order thickenings of algebraic spaces over $S$. Assume that $f$ is flat. Then the following are equivalent

1. $f'$ is flat and $X = Y \times _{Y'} X'$, and

2. the canonical map $f^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ is an isomorphism.

Proof. Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective étale morphism $U' \to X' \times _{Y'} V'$. Set $U = X \times _{X'} U'$ and $V = Y \times _{Y'} V'$. According to our definition of a flat morphism of algebraic spaces we see that the induced map $g : U \to V$ is a flat morphism of schemes and that $f'$ is flat if and only if the corresponding morphism $g' : U' \to V'$ is flat. Also, $X = Y \times _{Y'} X'$ if and only if $U = V \times _{V'} V'$. Finally, the map $f^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ is an isomorphism if and only if $g^*\mathcal{C}_{V/V'} \to \mathcal{C}_{U/U'}$ is an isomorphism. Hence the lemma follows from its analogue for morphisms of schemes, see More on Morphisms, Lemma 37.10.1. $\square$

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