Functoriality of internal hom of chain complexes

Question

The internal hom of chain complexes $[-,-]$ is supposed to form a bifunctor $$\operatorname{Ch}_\bullet(\mathsf{Mod}_R)^\mathrm{op} \times \operatorname{Ch}_\bullet(\mathsf{Mod}_R) \to \operatorname{Ch}_\bullet(\mathsf{Mod}_R).$$

The boundary map on $[X,Y]$ distributes over function composition as $d(\sigma \circ \tau ) = (d\sigma) \circ \tau + (-1)^{|\sigma|} \sigma \circ d\tau$.

Functoriality in the first variable says that a chain map $W_\bullet \xrightarrow{f} X_\bullet[k]$ should induce a chain map

$$[X_\bullet, Y_\bullet] \xrightarrow{f^*} [W_\bullet, Y_\bullet[k]] = [W_\bullet, Y_\bullet][k] \qquad \qquad \scriptscriptstyle{\dashleftarrow \text{ or should this be $-k$? Doesn't seem to matter}}$$

Now if $\sigma \in [X_\bullet, Y_\bullet]$ of degree $j$, then $$\require{cancel} d^{\operatorname{[W_\bullet,Y_{\bullet}][k]}}( \sigma \circ f) = (-1)^k d^{\operatorname{[W_\bullet,Y_{\bullet}]}}(\sigma \circ f) = (-1)^k \left( d\sigma \circ f + (-1)^{|\sigma|}\sigma \circ \cancel{df} \right) = (-1)^k d\sigma \circ f$$ $$ \neq d \sigma \circ f$$ So it seems not to be a chain map. Where am I going wrong?

Answer 1

Functoriality in the first variable only says that for $k = 0$ (morphisms have degree $0$). If $f$ is a morphism of degree $k$ then it induces a map

$$[X[k], Y] \to [W, Y]$$

and there are some signs involved in moving this shift $[k]$ around.