Hochschild homology: change of ground ring Theorem 9.1.7 in Weibel’s homological algebra reads as follows (I will change the notation slightly):
Let $f \colon k \to \ell$ be a morphism of commutative rings. Denote $\otimes = \otimes_k$. Let $A$ be a $k$-algebra. Consider $\ell \otimes A$: it is an $\ell$-algebra where $\ell$ acts on the left.
Let $Q$ be an $(\ell \otimes A)$-bimodule. In particular, it is an $A$-bimodule by restriction of scalars. Then:

$\operatorname{HH}^\ell_*(\ell \otimes A, Q) \cong \operatorname{HH}^k_*(A, Q)$.

He proves it as follows: consider the canonical $k$-isomorphism
$$
  Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell \, n} \cong Q \otimes A^{\otimes n} \,.
$$
He claims this yields an isomorphism of the Hochschild complexes, but I fail to see how it commutes with the last face map. So let’s look at it in detail. The isomorphism is
$$
  q \otimes((\lambda_1 \otimes a_1) \otimes \dotsb \otimes (\lambda_n \otimes a_n)) \mapsto (q \cdot (\lambda_1 \dotsm \lambda_n \otimes 1)) \otimes a_1\otimes \dots \otimes a_n.
$$
We need to check that the following square commutes.
$$
  \require{AMScd}
  \begin{CD}
    Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell \, n}
    @>>>
    Q \otimes A^{\otimes n}
    \\
    @VVV
    {}
    @VVV
    \\
    Q \otimes_\ell (\ell \otimes A)^{\otimes_\ell \, (n-1)}
    @>>>
    Q \otimes A^{\otimes (n-1)}
  \end{CD}
$$
But I believe that going first horizontally and then vertically makes $q\otimes((\lambda_1\otimes a_1)\otimes \dots \otimes (\lambda_n \otimes a_n))$ land in
$$
  (1\otimes a_n) \cdot q \cdot (\lambda_1 \cdots \lambda_n \otimes 1) \otimes a_2\otimes \dots \otimes a_n
$$
and going in the other direction makes it land in
$$
  (\lambda_n \otimes a_n) \cdot q \cdot (\lambda_1 \cdots \lambda_{n-1} \otimes 1) \otimes a_2 \otimes \dots \otimes a_n
$$
and I don’t see why these two should be equal. If the bimodule $Q$ were symmetric, this would be true, but why would it be true in general?
So where is my mistake?
 A: I think the question is what Weibel means by "$R_\ell$-$R_\ell$-bimodule" in his Theorem 9.1.7. When algebraists talk of an "$\left(A,B\right)$-bimodule" (Weibel writes "$A$-$B$-bimodule"), they rarely mean literally an additive group endowed with a left $A$-module structure and a right $B$-module structure satisfying the "associativity law"
(1) $\left(ac\right)b=a\left(cb\right)$ for all $a \in A$, $b \in B$ and $c \in C$.
Usually, what they mean is a relative version of this notion. Namely, they usually work over a commutative "base ring" $k$, for which both $A$ and $B$ are $k$-algebras. Then, they define an $\left(A,B\right)$-bimodule to mean a $k$-module endowed with a $k$-bilinear left $A$-module structure and a $k$-bilinear right $B$-module structure satisfying the "associativity law" (1). Strictly speaking, it is a bad idea to suppress the $k$ from the notation, as it leads to confusion (see, e.g., math.stackexchange #889130), but in practice it works most of the time when the $k$ is really clear from the context. (I personally prefer to keep the $k$ explicit and speak of "$\left(A,B\right)_k$-bimodules".)
Now, I think that when Weibel speaks of "$R_\ell$-$R_\ell$-bimodule", he wants $\ell$ (not $k$) to be the common base ring. If that is so, then your diagram commutes. This is less clear in your restatement of the theorem, because the tensor product $\ell \otimes A$ does not suggest a full base change as strongly as the notation $R_\ell$. But of course, the right solution is not to rely on suggestion, and clarify the notation.
