Let $C$ be a (small) category. One can form the free groupoid $GC$ of $C$ which is the left adjoint construction to the inclusion functor $\mathrm{Groupoid}\rightarrow\mathrm{Category}$. Is then $C$ always homotopy equivalent to $GC$? In other words, are the spaces $BC$ and $B\pi BC$ homotopy equivalent where $B\underline{}$ is the classifying space and $\pi\underline{}$ is the fundamental groupoid?
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asked Jun 7, 2013 at 14:38
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1$\begingroup$ Which adjoint? The use of the word "free" suggests you mean the left adjoint, but in fact there is also a right adjoint (maximal subgroupoid). $\endgroup$– Zhen LinCommented Jun 7, 2013 at 16:13
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$\begingroup$ Yes the word "free" pins it down: It's the left adjoint. In other words, you localize the category at all morphisms. I also changed the word "forgetful" to "inclusion". $\endgroup$– Werner ThumannCommented Jun 7, 2013 at 18:03
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No. There is a monoid with trivial group image whose classifying space is a sphere. See Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG?
Basically take the idempotent semigroup with elements $(a,b)$ with a,b either 0 or 1 and multiplication is (a,b)(c,d)=(a,d). Next add an identity. Clearly the fundamental group of this idempotent monoid is trivial. It is known to have classifying space homotopic to a 2-sphere.
answered Jun 7, 2013 at 14:44
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$\begingroup$ Thank you for your answer. Do you know whether the Ore condition on $C$ is enough for the statement to be true (you mentioned in the other question that it's true in the case of one object categories). Furthermore, do you even know of an easy to verify condition on $C$ which is equivalent to $C\rightarrow GC$ being an homotopy equivalence? $\endgroup$ Commented Jun 7, 2013 at 15:06
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$\begingroup$ Ore condition is enough for it to be true. I don't think there is an easy condition in general. When C is a cancellative monoid, the whole game boils down to showing that the homology of the monoid with group ring coefficients vanishes in dimension 2 and up if memory serves. The above question and its references discuss this. $\endgroup$ Commented Jun 7, 2013 at 19:55
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1$\begingroup$ In fact, you can realize any homotopy type as a nerve of a category. Just take a closed under intersections cover $U$ of topological space $X$ by contractible subspaces. Then the nerve of corresponding poset is equivalent to $X$. On the other hand, groupoids can model only homotopy 1-types, i.e. spaces with $\pi_i (X)=0$ for $i>1$. To get unrestricted equivalence, one needs to consider higher categories and $\infty$-groupoids. $\endgroup$ Commented Jun 7, 2013 at 21:43
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$\begingroup$ Dusa McDuff in fact proved every CW complex is homotopy equivalent to the classifying space of a monoid. $\endgroup$ Commented Jun 8, 2013 at 7:21
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$\begingroup$ Minor nitpick: McDuff's result is (obviously) for connected CW complexes, @BenjaminSteinberg. $\endgroup$ Commented Jun 27, 2013 at 14:42