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McDuff showed that every connected homotopy type can be realized as the classifying space of a discrete monoid, but the monoid she constructs has lots of idempotents.

Question: Which homotopy types are realized as the classifying space of a right-cancellative discrete monoid?

In the commutative case, my guess would be that $BM \simeq B(M[M^{-1}])$, so that the classifying space is aspherical. But I'm less confident that this happens in the noncommutative case.

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  • $\begingroup$ The category of elements of a semi-simplicial sets is left cancelative. So I don't know for monoid, but at least one can represent any homotopy type as the realization of a left (or right) cancelative category by taking the category of elements (or its opposite) of any semi-simplicial sets representing it. $\endgroup$ – Simon Henry Aug 22 '19 at 21:51
  • $\begingroup$ Indeed, by subdivision one can represent any homotopy type as the classifying space of a poset, which is both left and right cancellative. But far from being a monoid. $\endgroup$ – Tim Campion Aug 23 '19 at 0:26
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By universal properties, we have that $BM$ is the classifying space of the homotopy localization $B(M[M^{-1}]^h)$. Thus $BM$ is aspherical if and only if the homotopy localization is discrete.

Further, Dwyer-Kan showed that if $(M,W)$ admits a calculus of fractions, then the homotopy localization agrees with the ordinary localization. When $M$ is cancellative, $(M,M)$ admits a calculus of fractions if and only if it satisfies the Ore condition: $$\forall m_1, m_2 \in M, \exists n_1, n_2 \in M, ~ n_1 m_1 = n_2 m_2.$$ So in this case $M[M^{-1}]^h \simeq M[M^{-1}]$ and the homotopy localization is aspherical. In general, I do not know what happens.

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  • $\begingroup$ Thanks! I believe that every cancellative commutative monoid admits a calculus of fractions, but I'm not sure I believe this in the noncommutative case; do you have a reference? I don't see why the Ore condition should be satisfied. And for instance, having a calculus of fractions should probably imply that the monoid is embeddable in a group, which is not the case for all cancellative monoids, as I learned here. $\endgroup$ – Tim Campion Aug 22 '19 at 15:00
  • $\begingroup$ @TimCampion I think you're right, I was misreading the Ore condition on the nLab page, in such a way that I thought it was vacuous for the whole monoid. $\endgroup$ – Phil Tosteson Aug 22 '19 at 15:42

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