# When does nerving the groupoid completion of a category give a weak equivalence?

In an answer to another of my questions, Spice the Bird explains that for any monoid $M$, the map $NM\to NK(M)$ is a weak equivalence. Here $N:{\mathsf{Cat}}\to {\mathsf{sSet}}$ is the nerve functor and $K:{\mathsf{Mon}}\to{\mathsf{Grp}}$ is left adjoint to the forgetful functor ${\mathsf{Grp}}\to{\mathsf{Mon}}$. One views the monoid $M$ as a category with one object. The group $K(M)$ is often called the group completion of the monoid $M$.

More generally, for which small categories $C$ is the map $NC\to NK(C)$ is a weak equivalence? Here $K:{\mathsf{Cat}}\to{\mathsf{Grpd}}$ is left adjoint to the forgetful functor ${\mathsf{Grpd}}\to{\mathsf{Cat}}$. Similarly we can call the groupoid $K(C)$ the groupoid completion of the category $C$.

Since a groupoid is eqvuivalent as categories to a group, hence a necessary condition is that $NC$ is weakly equivalent to a $K(\pi,1)$. However this condition is not sufficient. For a counter example consider the category $C$ that looks like this $\bullet\rightrightarrows \bullet$. Here $NC$ is weakly contractible. On the other hand, the groupoid $K(C)$ is equivalent to the natural numbers ${\mathbb{N}}$ as categories, hence $NK(C)\simeq N{\mathbb{N}}$ which is weakly equivalent to $N{\mathbb{Z}}\simeq S^1$. Since $S^1$ has nontrivial fundamental group, therefore $NK(C)$ cannot be weakly equivalent to $NC$.

P.S. Is there a standard notation for the left adjoint to the forgetful functor ${\mathsf{Grpd}}\to{\mathsf{Cat}}$? I used $K$ to denote this in the above at the risk of confusion with the Grothendieck group or the $K$-theory ring.

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I don't understand your example. Isn't the nerve of $\bullet \rightrightarrows \bullet$ also weakly equivalent to a circle? – Charles Rezk Apr 21 '12 at 4:48
It is not true every monoid is weakly equivalent to its group completion. You need some Ore type condition. Also every simplicial complex is homeomorphic to the classifying space of its face lattice. So you don't always get an EM space. – Benjamin Steinberg Apr 21 '12 at 11:06
The answer that Gao 2Man references in the question explains why the nerves of cancellable monoids are equivalent to their group completions. – Dan Ramras Apr 21 '12 at 19:09
I don't really believe this works for cancellative monoids in general. I think you need some extra conditions. – Benjamin Steinberg Apr 21 '12 at 22:49