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**, J. Pure Appl. Algebra 175, No.1-3, 207-222 (2002). ZBL1015.18008.

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 $$L(K) = \varinjlim_{\Delta^n\to K} \Delta^n_{\text{$\infty$-grpd}}? $$