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!



Can one bypass the geometric realization in the definition of algebraic $K$-theory?

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



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


