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:

  1. 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 $S^2$.

  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.)

  3. 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.)

  4. 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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    $\begingroup$ If the monoid satisfies an Ore condition then the classifying spaces are equivalent. $\endgroup$ Commented Feb 18, 2013 at 20:18
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    $\begingroup$ Yes Ore=calculus of fractions. $\endgroup$ Commented Feb 18, 2013 at 21:10
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    $\begingroup$ Spice the bird seems to answer that no examples exist in his/her answer to mathoverflow.net/questions/94017/…. I never completely understood his/her answer so would be grateful if someone could. $\endgroup$ Commented Feb 18, 2013 at 22:28
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    $\begingroup$ In Fiedorowicz, Z. Classifying spaces of topological monoids and categories. Amer. J. Math. 106 (1984), no. 2, 301–350 it is shown that the natural map $BM\to BG$ is a homotopy equivalence if and ony if $H_n(M,\mathbb ZG)=0$ for $n\geq 1$. $\endgroup$ Commented Feb 19, 2013 at 3:14
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    $\begingroup$ If I understood correctly the classifying space of the slice category is the universal cover of BM. So it is contractible iff BM is a K(G,1). The homology of this slice category is the homology of M with coefficients in ZG. In particular BM is homotopy equivalent to BG iff the canonical map is an equivalence. $\endgroup$ Commented Feb 19, 2013 at 12:23


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