The inclusion $I\colon \mathbf{Grpd}\hookrightarrow\mathbf{Cat}$ of groupoids into categories has both a left and a right adjoint $L,R\colon \mathbf{Cat}\to \mathbf{Grpd}$, with $R(C)$ being largest groupoid contained in $C$ and $L(C) = C[C^{-1}]$ being $C$ with all morphisms brutally inverted. Going into $\infty$-category theory, we may replace $\mathbf{Cat}$ by weak Kan complexes (AKA quasi-categories) and $\mathbf{Grpd}$ by Kan complexes ($\infty$-groupoids). The inclusion $I\colon \mathbf{Kan}\hookrightarrow\mathbf{WKan}$ still has a right adjoint, given by taking largest contained Kan complex, as documented in Corollary 1.5 in *Joyal, A.*, [**Quasi-categories and Kan complexes**](http://dx.doi.org/10.1016/S0022-4049(02)00135-4), J. Pure Appl. Algebra 175, No.1-3, 207-222 (2002). [ZBL1015.18008](https://zbmath.org/?q=an:1015.18008).


However, the question now is: **Does it also have a left adjoint (possibly in some higher sense)?** This would have to be something like taking a weak Kan complex and brutally inverting all $1$-morphisms. Does this make sense in some way, possibly non-canonically?


**If no**, does there at least exist some $\infty$-groupoid analogue (“$\Delta^n\_{\text{$\infty$-grpd}}$”) of the standard simplex $\Delta^n\in\mathbf{SSet}$, behaving a bit like $L(\Delta^n)$, something like free Kan complex on $n+1$ objects? And in this case, why does this construction not solve my first problem by putting $$L(K) = \varinjlim\_{\Delta^n\to K} \Delta^n\_{\text{$\infty$-grpd}}? $$



Left adjoint of $I\colon \mathrm{Kan}\hookrightarrow\mathrm{WeakKan}$?
The inclusion of $\infty$-groupoids into $(\infty,1)$-categories certainly has a left adjoint. The functor $\mathbf{Kan} \hookrightarrow \mathbf{WKan}$ does not have a left adjoint, but the higher left adjoint can be modeled by any [fibrant replacement functor](https://ncatlab.org/nlab/show/Kan+fibrant+replacement).


We also can define $\Delta^n\_{\mathrm{\infty-grpd}}$ as the nerve of the terminal groupoid on $n$ objects. Your last formula doest not work since $L(K)$ may not be Kan. This happens when $K = \Lambda^2\_0$ for example.


