Left vs right projective resolutions and homology of monoids

Question

Let me use the ad hoc notation $\mathbb Z^l$ and $\mathbb Z^r$ to distinguish between left and right modules. These are trivial modules.

The homology of a (discrete) finite monoid $M$ with coefficient the trivial left $\mathbb Z M$-module $\mathbb Z^l$ can be defined as $H_i(M;\mathbb Z^l) := {\rm Tor}_i^{\mathbb Z M} (\mathbb Z^r, \mathbb Z^l)$. That is, one takes a resolution of $\mathbb Z^r$ by projective right $\mathbb Z M$-modules, applies $-\underset{\mathbb ZM}{\otimes}\mathbb Z^l$, and then takes the homology of the resulting chain complex.

One can also replace $M$ by $M^{\rm op}$ and calculate the homology of the latter. (This amounts to a certain switch of "left" and "right," but this way of thinking could be confusing.) In the case of groups, this produces no effect, thanks to the isomorphism $G^{\rm op} \cong G$ given by $g\mapsto g^{-1}$. For monoids, one expects $H_i(M^{\rm op};\mathbb Z^l)$ to be in general different from $H_i(M;\mathbb Z^l)$.

My questions:

1. Is it somehow true that $H_i(M^{\rm op};\mathbb Z^l) \cong H_i(M;\mathbb Z^l)$ for all finite monoids? For all monoids?

2. If the answer to any of the above is negative, could you provide a counterexample? If there are convenient ways to characterize when we have $H_i(M^{\rm op};\mathbb Z^l) \cong H_i(M;\mathbb Z^l)$, that would be really awesome!

I have computed the first few homology groups with coefficient $\mathbb Z^l$ of all monoids of order $\leq 7$, making use of the GAP package "smallsemi." It appears that for at least the majority (possibly all -- I only took a quick look) of these monoids, we have $H_i(M^{\rm op};\mathbb Z^l) \cong H_i(M;\mathbb Z^l)$, unless I have made some really silly mistake.

Among the monoids I considered are a large number of non-self-dual monoids (i.e. not isomorphic to the opposite monoids), which are necessarily non-commutative. Their homology groups thus do not necessarily agree with those of the group completions (see e.g. Fiedorowicz), and we cannot in general reduce the question whether $H_i(M^{\rm op};\mathbb Z^l)$ agrees with $H_i(M;\mathbb Z^l)$ to the case of groups.

Answer 1

This answer is due to Ammar Husain.

It is well-known that $H_i(M;\mathbb Z^l) \cong H_i(BM; \mathbb Z^l)$, where $BM$ is the (fat or usual) geometric realization of the nerve $BM_\bullet$ of the category $M$. Via the set map \begin{eqnarray} BM_n &\rightarrow& (BM^{\rm op})_n \\ (m_n, \ldots, m_1) &\mapsto& (m_1, \ldots, m_n) \end{eqnarray} the face map $d_i:BM_n\rightarrow BM_{n-1}$ becomes $d_{n-i}: (BM^{\rm op})_n \rightarrow (BM^{\rm op})_{n-1}$, and the degeneracy $s_i:BM_n\rightarrow BM_{n+1}$ becomes $s_{n-i}: (BM^{\rm op})_n \rightarrow (BM^{\rm op})_{n+1}$. Then it is clear that $BM_\bullet$ and $(BM^{\rm op})_\bullet$ have the same geometric realization.

At the level of ${\rm Tor}$ groups, what is happening is that the chain complex (obtained after applying $-\underset{\mathbb Z M}{\otimes} \mathbb Z$) \begin{equation} \cdots \xrightarrow{d} \mathbb Z[M^{{\rm op}~n}] \xrightarrow{d} \mathbb Z[M^{{\rm op}~n-1}] \xrightarrow{d} \cdots \xrightarrow{d} \mathbb Z[M^{{\rm op}~0}] \rightarrow 0 \end{equation} for $M^{\rm op}$ is isomorphic to \begin{equation} \cdots \xrightarrow{d} \mathbb Z[M^n] \xrightarrow{d} \mathbb Z[M^{n-1}] \xrightarrow{d} \cdots \xrightarrow{d} \mathbb Z[M^0] \rightarrow 0 \end{equation} but with $d: \mathbb Z[M^n] \rightarrow \mathbb Z[M^{n-1}]$ replaced by $(-1)^n d$. This obviously does not change the homology groups.