Let $A$ be an associative algebra, $Z(A)$ the centre of $A$, $C^n(A, A) = \mbox{Hom}_{A^e}(A^{\otimes n +2}, A)$ the $n$-th component of the Hochschild cochain complex, $b:C^n(A,A) \to C^{n+1}(A, A)$ the Hochschild coboundary map and $HH^n(A)$ the $n$-th Hochschild cohomology of $A$ (with coefficients in $A$). Define the action of $Z(A)$ on $C^n(A, A)$ by
$$(a\cdot f)(a_1\otimes \dots\otimes a_{n+2}) = af(a_1\otimes \dots\otimes a_{n+2}) $$
For $a\in Z(A)$ and $f\in C^n(A,A)$. Now, I'm reading Hochschild Cohomology for Algebras by Sarah Witherspoon, and in the section that discusses this action she writes
this action commutes with the differentials on the bar complex, inducing an action of $Z(A)$ on $HH^n(A)$ under which $HH^n(A)$ becomes a $Z(A)$-module.
(I'm assuming that the use of the term "bar complex" here is just for ease since $C^*(A, A)$ is technically the cochain complex you get after applying the functor $\mbox{Hom}_{A^e}(-, A)$ to the bar complex, rather than the bar complex itself.)
My question is, why does the fact that this action commutes with $b$ mean that it induces an action of $Z(A)$ on $HH^n(A)$?
My initial thoughts about this is that since $a\in Z(A)$ then
$$af(a_1\otimes \dots\otimes a_{n+2}) = f(a_1\otimes \dots\otimes a_{n+2})a$$
and since it also commutes with the differential then $a\cdot [f] := [af] = [fa]$. But if $a$ already commutes with $f$ then doesn't that automatically make the equivalence classes $[af]$ and $[fa]$ the same, regardless of whether $a$ commutes with $b$? So why is this commutativity with $b$ necessary? I feel like I'm missing something obvious here!
I've figured out where my confusion was coming from.
First off I misunderstood the term "commutes" in this context. It's referring to the fact that
$$(a\cdot b(f))(a_1\otimes\dots\otimes a_{n+2}) = b(a\cdot f)(a_1\otimes\dots\otimes a_{n+2})$$
Therefore, if we have $[f] = [g] \in HH^n(A)$, then this implies that there is some map $h$ such that
$$f - g = b(h)$$
and so applying our action to this expression we get
$$a\cdot f - a\cdot g = a\cdot (f- g) = a\cdot b(h) = b(a\cdot h) $$
and therefore $[a\cdot f] = [a\cdot g]$. So the commutativity of the action with the differential means that the action
$$a\cdot [f] = [a\cdot f] $$
of $Z(A)$ on $HH^n(A)$ is well defined. And then showing that $HH^n(A)$ is a $Z(A)$-module is an easy exercise of checking the definition.