Lemma 10.134.14. Let $R$ be a ring. Let $A_1 \to A_0$, and $B_1 \to B_0$ be two term complexes. Suppose that there exist morphisms of complexes $\varphi : A_\bullet \to B_\bullet$ and $\psi : B_\bullet \to A_\bullet$ such that $\varphi \circ \psi$ and $\psi \circ \varphi$ are homotopic to the identity maps. Then $A_1 \oplus B_0 \cong B_1 \oplus A_0$ as $R$-modules.

Proof. Choose a map $h : A_0 \to A_1$ such that

$\text{id}_{A_1} - \psi _1 \circ \varphi _1 = h \circ d_ A \text{ and } \text{id}_{A_0} - \psi _0 \circ \varphi _0 = d_ A \circ h.$

Similarly, choose a map $h' : B_0 \to B_1$ such that

$\text{id}_{B_1} - \varphi _1 \circ \psi _1 = h' \circ d_ B \text{ and } \text{id}_{B_0} - \varphi _0 \circ \psi _0 = d_ B \circ h'.$

A trivial computation shows that

$\left( \begin{matrix} \text{id}_{A_1} & -h' \circ \psi _1 + h \circ \psi _0 \\ 0 & \text{id}_{B_0} \end{matrix} \right) = \left( \begin{matrix} \psi _1 & h \\ -d_ B & \varphi _0 \end{matrix} \right) \left( \begin{matrix} \varphi _1 & - h' \\ d_ A & \psi _0 \end{matrix} \right)$

This shows that both matrices on the right hand side are invertible and proves the lemma. $\square$

Comment #1721 by Yogesh More on

In the proof of tag 00S3, $h:A_0 \to B_1$ should be $A_0 \to A_1$, and similarly $h':B_0 \to B_1$. In the displayed equation asserting that that $h'$ is a homotopy, you have $h$ instead of $h'$.

In Tag 00S4 line 2 of the statement of the theorem says "Let $\alpha':P' \otimes_R R' \to S'$ be the induced presentation", the $P'$ should be a $P$.

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