Kerodon

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Warning 2.5.1.14 (The Koszul Sign Rule). Let $(C_{\ast }, \partial )$ and $( D_{\ast }, \partial )$ be chain complexes. There is a unique isomorphism of graded abelian groups $\tau : C_{\ast } \boxtimes D_{\ast } \rightarrow D_{\ast } \boxtimes C_{\ast }$ satisfying $\tau (x \boxtimes y) = y \boxtimes x$ for all $x \in C_{m}$, $y \in C_{n}$. Beware that $\tau $ is usually not a chain map: we have

\[ \partial \tau (x \boxtimes y) = \partial (y \boxtimes x) = (\partial y \boxtimes x) + (-1)^{n} (y \boxtimes \partial x) \]
\[ \tau ( \partial (x \boxtimes y) ) = \tau ( (\partial x \boxtimes y) + (-1)^{m} (x \boxtimes \partial y) ) = (-1)^{m} (\partial y \boxtimes x) + (\partial x \boxtimes y). \]

This can be remedied by modifying the isomorphism $\tau $: there is another isomorphism of graded abelian groups

\[ \sigma : C_{\ast } \boxtimes D_{\ast } \simeq D_{\ast } \boxtimes C_{\ast } \quad \quad \sigma (x \boxtimes y) = (-1)^{mn} (y \boxtimes x). \]

The isomorphism of $\sigma $ is a chain map (hence an isomorphism of chain complexes) by virtue of the calculation

\begin{eqnarray*} \partial \sigma (x \boxtimes y) & = & \partial ( (-1)^{mn} y \boxtimes x) \\ & = & (-1)^{mn} (\partial y \boxtimes x) + (-1)^{mn+n} (y \boxtimes \partial x) \\ & = & (-1)^{n} \sigma (x \boxtimes \partial y) + \sigma (\partial x \boxtimes y) \\ & = & \sigma ( \partial ( x \boxtimes y) ). \end{eqnarray*}