Lemma 46.3.5. Let $A$ be a ring. Let $F$ be an adequate functor on $\textit{Alg}_ A$. If $B \to B'$ is flat, then $F(B) \otimes _ B B' \to F(B')$ is an isomorphism.
Proof. Choose an exact sequence $0 \to F \to \underline{M} \to \underline{N}$. This gives the diagram
\[ \xymatrix{ 0 \ar[r] & F(B) \otimes _ B B' \ar[r] \ar[d] & (M \otimes _ A B)\otimes _ B B' \ar[r] \ar[d] & (N \otimes _ A B)\otimes _ B B' \ar[d] \\ 0 \ar[r] & F(B') \ar[r] & M \otimes _ A B' \ar[r] & N \otimes _ A B' } \]
where the rows are exact (the top one because $B \to B'$ is flat). Since the right two vertical arrows are isomorphisms, so is the left one. $\square$
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