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