• There is a canonical identification

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet )[a - b] \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a])$

which uses signs. It is defined as the map whose corresponding shifted map

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a])[b - a]$

uses the sign $(-1)^{nb}$ on the module $\mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ p)$ with $p + q = n$. Namely, if $f \in \mathop{\mathrm{Hom}}\nolimits ^ n(M^\bullet , K^\bullet )$ then

$d(f) = d_ K \circ f - (-1)^ n f \circ d_ M$

on the source, whereas on the target $f$ lies in $\left(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a])[b - a]\right)^ n = \mathop{\mathrm{Hom}}\nolimits ^{n + b -a}(M^\bullet [b], K^\bullet [a])$ and hence we get

\begin{align*} d(f) & = (-1)^{b - a} \left(d_{K[a]} \circ f - (-1)^{n + b - a} f \circ d_{M[b]}\right) \\ & = (-1)^{b - a} \left((-1)^ a d_ K \circ f - (-1)^{n + b - a} f \circ (-1)^ b d_ M \right) \\ & = (-1)^ b d_ K \circ f - (-1)^{n + b} f \circ d_ M \end{align*}

and one sees that the chosen sign of $(-1)^{nb}$ in degree $n$ produces a map of complexes for these differentials.

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