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_ Mon 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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