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I believe there is no good notion of homotopy groups for an arbitrary simplicial set $S$. However, when $S$ is fibrant - meaning that $S\to *$ is a fibration - there is a definition. The singular realization of a topological space is fibrant, and one recovers the homotopy groups this way. Given a small category $\mathcal{C}$, its nerve $N\mathcal{C}$ is a simplicial set which is almost never fibrant; it is, though, whenever $\mathcal{C}$ is a groupoid.

Given an exact category $\mathcal{E}$, Quillen defines another category $Q(\mathcal{E})$ whose objects are the same as $\mathcal{E}$ but morphisms are given by "roofs" $\bullet \twoheadleftarrow \bullet \hookrightarrow \bullet$ where the epi and the mono are parts of exact sequences in $\mathcal{E}$. Composition is given by pullbacks, and is well-defined due to the axioms of exact categories.

Then, Quillen defines the $K$-group $K_i(\mathcal{E})$ as the $(i+1)$th homotopy group of the topological realization of the simplicial set given by the nerve $N\mathcal{E}$ of $\mathcal{E}$.

I am curious: since one can already define homotopy groups of simplicial sets, is there a way to bypass the topological realization of $N\mathcal{E}$?

Since the category $\mathcal{E}$ is almost never a groupoid (although all morphisms are mono), I believe that the simplicial set $N\mathcal{E}$ is not fibrant. Yet, can one replace $Q(\mathcal{E})$ by some groupoid $G(\mathcal{E})$ functorially attached to $\mathcal{E}$ and define $K_{i}(\mathcal{E})$ directly as $\pi_{i+1}(NG(\mathcal{E}))$?

Sorry if this all well-known. Many thanks!

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    $\begingroup$ You don't need topological spaces to construct a fibrant replacement. For instance you can use Kan's Ex infinity functor: ncatlab.org/nlab/show/Kan+fibrant+replacement $\endgroup$
    – Phil Tosteson
    Commented Aug 4, 2023 at 13:30
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    $\begingroup$ In what way is $\mathcal{Q}(\mathcal{E})$ intermediate? That is the central construction in this method of defining $K$-theory. $\endgroup$
    – Connor Malin
    Commented Aug 4, 2023 at 14:23
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    $\begingroup$ It can't be the case that K-theory factors as $\pi_* \circ \mathrm{N} \circ F$ where $F:\mathrm{ExactCat} \rightarrow \mathrm{Grpd}$ because the higher homotopy groups of the nerve of a groupoid are all trivial, and in general higher $K$-theory is nontrivial. $\endgroup$
    – Connor Malin
    Commented Aug 4, 2023 at 16:14
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    $\begingroup$ To be clear, Quillen's definition is that the K-theory groups of $\mathcal{E}$ are the homotopy groups of $NQ(\mathcal{E})$, not $N(\mathcal{E})$ (so this is a typo in the question). $\endgroup$
    – Dan Ramras
    Commented Aug 5, 2023 at 2:22
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    $\begingroup$ One answer to the question in the title is provided by work of Grayson (ams.org/journals/jams/2012-25-04/S0894-0347-2012-00743-7/…) where he gives presentations of the K-theory groups in terms of combinatorial data about $\mathcal{E}$. See also more recent work of Kasprowski and Winges (arXiv:1705.09116, arXiv:1805.06175) $\endgroup$
    – Dan Ramras
    Commented Aug 5, 2023 at 2:24

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I believe there is no good notion of homotopy groups for an arbitrary simplicial set S.

It depends on what “good” means. Kan's original definition works for arbitrary pointed simplicial sets: $$\def\Exi{\mathop{\sf Ex^∞}}π_k(S,*):=[S^k,\Exi S],$$ where $\Exi$ was defined by Kan in 1950s and $S^k$ can be taken to be any simplicial model for the $n$-sphere, e.g., $S^k=Δ^k/∂Δ^k$. (A modern exposition is found in Goerss–Jardine, Section III.4.) If “good” means usable in practice, there are many situations where the above definition is good.

is there a way to bypass the topological realization of NE?

Yes, we can use the above definition of homotopy groups.

Yet, can one replace Q(E) by some groupoid G(E) functorially attached to E and define Ki(E) directly as πi+1(NG(E))?

No, because the higher homotopy groups of the nerve of any groupoid necessarily vanish: $π_k(NG,*)≅0$ for $k>1$ and any basepoint $*$. (That is to say, the nerve is a disjoint union of Eilenberg–MacLane spaces.)

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  • $\begingroup$ Many thanks! This is helpful. I also got me a copy of Goerss-Jardine! $\endgroup$
    – Stabilo
    Commented Aug 5, 2023 at 8:48
  • $\begingroup$ A curiosity: modeling $\infty$-categories by weak Kan complexes, thanks to the definition you mention, one could define homotopy groups of a $\infty$-category $\mathcal{C}$ (say, pointed by an initial object). Is there an interpretation of these, e.g. when $\mathcal{C}$ is the nerve of an ordinary category? $\endgroup$
    – Stabilo
    Commented Aug 5, 2023 at 9:07
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    $\begingroup$ @Stabilo: Of course: $\def\Exi{\mathop{\sf Ex^∞}}\def\cC{{\cal C}}\Exi \cC$ can be shown to present the ∞-groupoid $\cC[\cC^{-1}]$, i.e., $\cC$ with all of its morphisms inverted up to homotopy. Even if $\cC$ is an ordinary category, $\cC[\cC^{-1}]$ is typically an ∞-groupoid, not a 1-groupoid. The $n$th homotopy group of this ∞-groupoid is precisely the group of isomorphism classes of n-morphisms in $\cC[\cC^{-1}]$ with source and target being identities on the basepoint object and the group structure being given by composition in $\cC[\cC^{-1}]$. $\endgroup$
    – Dmitri Pavlov
    Commented Aug 5, 2023 at 20:39
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    $\begingroup$ @Stabilo: Using a concrete description of $\def\Exi{\mathop{\sf Ex^∞}}\def\cC{{\cal C}}\Exi \cC$, you can interpret an element in the $n$th homotopy group of $\Exi \cC$ as a diagram in $\cC$ whose indexing category is given by the subdivided $n$-sphere (which is always a poset if you subdivide two or more times). $\endgroup$
    – Dmitri Pavlov
    Commented Aug 5, 2023 at 20:43

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