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What is an example of a simplicial set $S$ such that its homotopy category $hS$ is a groupoid, but such that $S$ is not an $\infty$-category?

I know that if $S$ is an $\infty$-category, then $S$ is a Kan complex if and only if $hS$ is a groupoid. My question is about what happens if $S$ is not an $\infty$-category.

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  • $\begingroup$ What is the definition of the homotopy category of a simplicial set which is not an $\infty$-category? I would have just defined it to be the homotopy category of a Joyal-fibrant replacement... $\endgroup$
    – Tim Campion
    Commented Dec 4, 2021 at 19:06
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    $\begingroup$ @TimCampion I would assume that it is the free category generated by the 1-simplices under the relations given by the 2-simplices. I suspect this corresponds to the homotopy category of a Joyal-fibrant replacement though (since this construction looks invariant under categorical equivalences) $\endgroup$
    – Denis Nardin
    Commented Dec 4, 2021 at 21:50

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Since the homotopy category depends only on the 2-skeleton of the simplicial set, the easiest thing to do is to take the 2-skeleton of an ∞-groupoid. For example, let $E$ be the nerve of the contractible groupoid on two objects. This has as a set of $n$-simplices the set $\{a,b\}^{[n]}$ of functions (of sets) from $[n]=\{0<...<n\}$ to the set $\{a,b\}$ (therefore the sequences of $n+1$ letters $a$ or $b$).

Then the 2-skeleton $\operatorname{sk}_2E$ of this simplicial set is just the subset of those functions $[n]\to \{a,b\}$ that factor via a non-decreasing map $[n]\to [2]$. This is not an $\infty$-category, however its homotopy category is the same as the homotopy category of $E$, and so it is exactly a contractible groupoid.

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