For a (discrete) monoid $M$, the classifying space $BM$ is the geometric realization of the nerve of the one object category whose hom-set is $M$. (This definition gives the usual classfiying space when $M$ is a group.) The group completion of $M$ can be constructed as the fundamental group of $BM$, and is characterized by the universal property that any monoid homomorphism from $M$ to a group factors uniquely through the group completion.

My question is whether there is an example of a monoid for which the canonical map to its group completion is injective, but for which this canonical map does not induce a homotopy equivalence of the classifying spaces.

As background here are some facts:

Classifying spaces of monoids produce all connected homotopy types! This is proved in Dusa McDuff's 1979 paper

*On the classifying spaces of discrete monoids*. For a neat concrete example, see Zbigniew Fiedorowicz's*A counterexample to a group completion conjecture of JC Moore*; it shows a specific 5 element monoid whose classifying space is homotopy equivalent to $S^2$.If $G$ is the group completion of a

*commutative*monoid $M$, the canonical map $BM \to BG$ is a homotopy equivalence; even if $M \to G$ is*not*injective. (This is easy to prove: think of $M \to G$ as a functor between one object categories and apply Quillen's Theorem A to it. There is only one slice category to check and using commutativity it is easy to see this category is filtered and thus contractible.)If $M$ is a free monoid and the free group $G$ is its completion, the map $BM \to BG$ is a homotopy equivalence. It fact, more generally, if $C$ is the free category on some directed graph $X$, the nerve of $C$ is homotopy equivalent to the geometric realization of $X$. This is proved in Dwyer and Kan's

*Simplical Localization of Categories*, proposition 2.9, but the proof is simple enough to sketch here: for each $k$, the inclusion of the $k$-skeletion of $NC$ into the $(k+1)$-skeleton is a weak homotopy equivalence (since you get the $(k+1)$-skeleton by filling in some horns); so the $1$-skeleton, $X$, is weakly equivalent to $NC$. (The claim for free monoids is the case where $X$ consists of a single vertex with some loops.)Even if a monoid has left and right cancellation the canonical map to its group completion might not be injective. Here's an example from Malcev's

*On the Immersion of an Algebraic Ring into a Field*: let $M$ be the*monoid*presented by $(a,b,c,d,x,y,u,v : ax=by, cx=dy, au=bv)$. Malcev shows that $M$ is cancellative, but that in $M$, $cu \neq dv$; in any group the relations listed for $M$ would imply that $cu=dv$.

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