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

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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.

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.

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  • $\begingroup$ What do you mean by the inclusion of $(\infty,1)$-categories into $\infty$-groupoids? The only inclusion I can think of goes in the other direction. $\endgroup$ – Gaussler Dec 11 '17 at 14:00
  • $\begingroup$ Yes, you're right. I've edited the answer. $\endgroup$ – Valery Isaev Dec 11 '17 at 14:05
  • $\begingroup$ What if you tried taking the colimit inside the category of Kan complexes? $\endgroup$ – Gaussler Dec 12 '17 at 7:31
  • $\begingroup$ @Gaussler The category of Kan complexes does not have all colimits. $\endgroup$ – Valery Isaev Dec 12 '17 at 9:27
  • $\begingroup$ Thanks for that information, I sought in vain in the literature for the answer to the question of cocompleteness. $\endgroup$ – Gaussler Dec 12 '17 at 9:32
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As Valery Isaev has pointed out, $\mathbf{Kan} \hookrightarrow \mathbf{WKan}$ does not have a left adjoint. However, instead of considering $\mathbf{Kan} \hookrightarrow \mathbf{WKan}$, you can consider the forgetful functor $\mathbf{sKan} \hookrightarrow \mathbf{sWKan}$ of (weak) Kan complexes with chosen Kan fillers and morphisms preserving these choices. (Don't google this, I've just made up the notation.) The two categories are locally presentable (they can be axiomatized as essentially algebraic theories), and every forgetful functor between such categories is a right adjoint. Because the corresponding left adjoint is given by repeated pushouts along the inclusion $\Lambda^k_n \rightarrow \Delta^n$, it follows that the unit of the adjunction is a fibrant replacement wrt. the model structure on simplicial sets that has the Kan complexes as fibrant objects.

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