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I'm learning the abstract definition of the $i$-th $K$-group $K_i(\mathcal C)$ for an exact category $\mathcal C$.

The definition is quite complicate, basically you start from an exact category $\mathcal C$ (by simplicity a particular full subcategory of an abelian category). Then you construct the Quillen category $\mathcal Q(\mathcal C)$ which has the same objects of $\mathcal C$ but with different morphisms. At this point you take the topological space $\mathcal B\mathcal Q(\mathcal C)$ which is the geometric realization of the nerve of the category $\mathcal Q(\mathcal C)$. Finally you define: $$K_i(\mathcal C)=\pi_{i+1}(\mathcal B\mathcal Q(\mathcal C))$$ I don't write any base point because this construction is independent from the base point (maybe $\mathcal B\mathcal Q(\mathcal C)$ is path connected?).


Now I want to ask some questions:

  • The construction of $K_i(\cdot)$ should be functorial in the following sense. If $F:\mathcal C\to\mathcal D$ is an exact functor between exact categories, then there is a homomorphism $F_\ast:K_i(\mathcal C)\to\mathcal K_i(\mathcal D)$. Do you know how is constructed (maybe also vaguely) the map $F_\ast$?
  • In general the homotopy groups are not abelian, but what about $K_i(\mathcal C)$? Is it abelian? I'm asking this because it seems that the $K$-groups associated to schemes (i.e. to the category of locally free sheaves) are abelian, but I don't understand the reason.
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  • $\pi_{i+1} (X)$ is always abelian for $i > 0$, and for $i = 0$ it is a result of Quillen from his original paper "Higher algebraic K-theory" (Theorem 1, p. 18) that verifies that the fundamental group of $BQ\mathcal{C}$ coincides with the usual Grothendieck group $K_0 (\mathcal{C})$, which is by definition abelian.

  • An exact functor $\mathcal{C} \to \mathcal{D}$ induces functorially morphisms $K_i (\mathcal{C}) \to K_i (\mathcal{D})$. This follows from functoriality of $\pi_{i+1}$, $B$, and $Q$. For the functoriality of $Q$, see p. 18 of Quillen's paper (the universal property of the Q-construction).

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