Proof of Corollary in localizing $\operatorname{Tor}_*$ in Weibel In Weibel's An introduction to homological algebra, one has the following lemma (p 72):
Let $R$ be a commutative ring.

Lemma 3.2.11 If $\mu : A \rightarrow A$ is multiplication by a central element $r \in R$, so are the induced maps $\mu_* : \operatorname{Tor}^R_n(A,B) \rightarrow \operatorname{Tor}^R_n(A,B)$ for all $n$ and $B$.

I would like to prove the following corollary:

Corollary 3.2.12 If $A$ is an $R/r$-module, then for every $R$-module $B$ the $R$-modules $\operatorname{Tor}_*^R(A,B)$ are actually $R/r$-modules, that is, annihilated by the ideal $rR$. 

$\textbf{My attempt:}$
We want to show that $rR$ also annihilates the $R$-modules $\operatorname{Tor}_*^R(A,B)$.
Since $\forall r' \in rR$, $r' = rc$ for some $c \in R$, it suffices to show that the induced maps $\mu_* : \operatorname{Tor}^R_n(A,B) \rightarrow \operatorname{Tor}^R_n(A,B)$ are the zero map for all $n$ and $B$, where $\mu : A \rightarrow A, a \mapsto ra$. Since multiplication by $r$ is an $R$-module chain map $\tilde{\mu} : P \rightarrow P$ where $P \rightarrow A$ is a projective resolution
$$\begin{matrix}
...\to&P_1&\xrightarrow{d_1} &P_0&\xrightarrow{d_0}&A&\rightarrow &0\\
&\downarrow\rlap{\scriptstyle\tilde{\mu}_1}&&\downarrow\rlap{\scriptstyle\tilde{\mu}_0}&&\downarrow\rlap{\scriptstyle\mu}\\
...\to &P_1&\xrightarrow{d_1} &P_0&\xrightarrow{d_0} &A&\rightarrow &0
\end{matrix} $$
As $\_ \otimes_R B$ is right exact, we have 
$$\begin{matrix}
...\to&P_1 \otimes B&\xrightarrow{d_1 \otimes B} &P_0\otimes B&\xrightarrow{d_0 \otimes B}&A\otimes B&\rightarrow &0\\
&\downarrow\rlap{\scriptstyle\tilde{\mu}_1\otimes B}&&\downarrow\rlap{\scriptstyle\tilde{\mu}_0\otimes B}&&\downarrow\rlap{\scriptstyle\mu\otimes B}\\
...\to &P_1\otimes B&\xrightarrow{d_1 \otimes B} &P_0\otimes B&\xrightarrow{d_0 \otimes B} &A\otimes B&\rightarrow &0
\end{matrix} $$
By assumption, $rR$ annihilates $A$, hence $\mu\otimes B : A\otimes B \to A\otimes B, a\otimes b \mapsto ra\otimes b = 0$ is the zero map. Since the
following diagram commutes
$$\begin{matrix}
P_0\otimes B&\xrightarrow{d_0 \otimes B}&A\otimes B\\
\downarrow\rlap{\scriptstyle\tilde{\mu}_0\otimes B}&&\downarrow\rlap{\scriptstyle\mu\otimes B}\\
P_0\otimes B&\xrightarrow{d_0 \otimes B} &A\otimes B
\end{matrix} $$
i.e. $(\mu\otimes B) \circ (d_0\otimes B )= (d_0\otimes B )\circ (\tilde{\mu}_0\otimes B)$ and $(d_0\otimes B )$ may not identically be the $0$ map, then this implies that $\tilde{\mu}_0\otimes B$ has to be the $0$ map. By induction, we can prove that $\tilde{\mu}_n\otimes B$ are $0$ maps for all $n$. Hence the induced map $\mu_*$ on the subquotient $Tor^R_n(A,B)$ of $P_n \otimes B$ is obviously the zero map for all $n$. Therefore, $Tor^R_n(A,B)$ is annihilated by $rR$. $\square$
Does the proof seem ok? Is there something missing in the proof? Any comments are greatly appreciated.
 A: So basically you are trying to prove "by hand" that the zero map $A\to A$ induces the zero map $Tor_n^R(A,B)\to Tor_n^R(A,B)$. This is automatic since the $Tor$ functors are additive, but let's assume we want to check it explicitly. Your proof has a few issues.
First, it's not a mistake per se, but there is no need for $\bullet \otimes_R B$ to be right exact to induce a map of chain complexes: any additive functor works (the only thing needed is that the zero map is sent to the zero map).

The first real issue is that you are not working with the correct complex. In general if $$\dots\to P_2\to P_1\to P_0\to A\to 0$$ is a projective resolution of $A$ and $F$ is a right exact functor (in this case $F(X)=X\otimes_R B$), the derived functors are defined by the homology of the complex $$\dots \to F(P_2)\to F(P_1)\to F(P_0)\to 0\quad (*)$$
and not $$\dots \to F(P_2)\to F(P_1)\to F(P_0)\to F(A)\to 0.$$
Note that since the projective resolution is by definition exact, and $F$ is right exact, then $F(P_1)\to F(P_0)\to F(A)\to 0$ is exact, but the complex $(*)$ is in general not exact at any point. In particular, $F(P_2)\to F(P_1)\to F(P_0)$ is not exact in general (it is exact exactly when $L_1F(A)=0$, so in this case $Tor_1^R(A,B)=0$).
Note that it is easy to recover $Tor_0^R(A,B)=A\otimes_R B$: by definition, $Tor_0^R(A,B)$ is the homology of $F(P_1)\to F(P_0)\to 0$, so it is the cokernel of $F(P_1)\to F(P_0)$. But since $F(P_1)\to F(P_0)\to F(A)\to 0$ is exact, the cokernel of $F(P_1)\to F(P_0)$ is naturally isomorphic to $F(A)$.

Another issue is your proof that $\tilde{\mu_0}\otimes B=0$. It is not necessarily the case: all you can deduce is that it has image in the kernel of $d_0\otimes B$. But the point is that even though it is not always the zero map, it always induces the zero map in the homology of the complex $(*)$.
The first way to see that is to use general properties: it is a basic fact of homological algebra that any choice of maps $\tilde{\mu_n}$, as long as they are compatible with the differentials, will induce the same map in the homology of $(*)$ (because two choices will always be chain homotopic). Since we can clearly choose $\tilde{\mu_n}=0$, any choice of $\tilde{\mu_n}$ must induce the zero map in homology.
If one wants to show everything by hand, we note that $\tilde{\mu_0}\otimes B$ has value in the kernel of $d_0\otimes B$; but by right exactess this is the image of $d_1\otimes B$, which is by definition the part of $P_0\otimes_R B$ which is sent to $0$ in the homology of the complex $(*)$ (so in $Tor_0^R(A,B)$).
Showing explicitly recursively that $\tilde{\mu_n}$ induces the zero map in homology for all $n$ is a little more tricky than you seemed to think in your answer, but it is not extremely difficult either. Just follow the proof that two choices of chain maps are homotopic, and adapt it to the case where one of the choices is the zero map.
A: We can compute $\mathrm{Tor}(A,B)$ starting from a projective resolution of $B$
$$ \cdots \to Q_2 \to Q_1 \to Q_0 \to B \to 0. $$
Tensoring with $A$ yields the complex
$$ \cdots \to A\otimes_R Q_2 \to A\otimes_R Q_1 \to A\otimes_R Q_0 \to 0 $$
whose homology groups are $\mathrm{Tor}^R_n(A,B)$. It is now clear that if $r\in R$ acts as zero on $A$, then $r$ acts as zero on each $A\otimes_RQ_n$, and so acts as zero on each $\mathrm{Tor}_n^R(A,B)$.
In fact, all this works in the noncommutative setting, so if $B$ is a left $R$-module and $A$ is an $S$-$R$-bimodule, then each $A\otimes_RQ_n$ is a left $S$-module, and so each $\mathrm{Tor}^R_n(A,B)$ is a left $S$-module.
