Let $C$ be a category. The *groupoid completion* of $C$ is the free groupoid on $C$, i.e. the category $C[C^{-1}]$ obtained by localizing at everything. Recall that the classifying space $\mathbf{B}C$ of $C$ is the geometric realization of the nerve of $C$. It's well-known that there is an equivalence of groupoids $C[C^{-1}] \simeq \Pi\_1(\mathbf{B}C)$ between the groupoid completion of $C$ and the fundamental groupoid of the classifying space of $C$.


Now let $\mathcal{C}$ be a topologically-enriched category. Now the groupoid completion $\mathcal{C}[C^{-1}]$ is a topological groupoid. There is also a classifying space $\mathbb{B}\mathcal{C}$, which is the geometric realization of the nerve of $\mathcal{C}$ (where the nerve of $\mathcal{C}$ is a simplicial space). But the fundamental groupoid $\Pi\_1(\mathbb{B}\mathcal{C})$ is discrete, so it is generally not equivalent to $\mathcal{C}[\mathcal{C}^{-1}]$.


**Question.**


Can we recover $\mathcal{C}[C^{-1}]$ up to equivalence of topological groupoids from $\mathbb{B}\mathcal{C}$ in some other way? More generally, what kind of homotopical significance does $\mathcal{C}[\mathcal{C}^{-1}]$ have? Does it fit into a sequence of topological groups analogous to the usual homotopy groups, and is there perhaps a natural model structure on topologically-enriched categories or simplicial spaces where the weak equivalences are the morphisms inverted by these functors?



Can the groupoid completion of a topological category be recovered from its classifying space?
I'm not sure how well-posed this question is, what exactly is the topological structure of $\mathcal{C}[\mathcal{C}^{-1}]$?


However, the question certainly makes good sense for simplicial categories and the answer is affirmative. Dwyer and Kan proved that if $\mathcal{C}$ is a cofibrant simplicial category, then $\mathcal{C}[\mathcal{C}^{-1}](X, Y)$ is the loop space of $N \mathcal{C}$ (at $X$). See the proof of Proposition 9.3 in *Simplicial Localizations of Categories* (this part of their results really goes back to Segal and his delooping machine).


