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. $\square$
Comments (2)
Comment #1721 by Yogesh More on
Comment #1760 by Johan on
There are also: