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The Eilenberg-Mazur swindle shows that the Grothendieck group of an additive category with countable coproducts is trivial. The strategy is to observe that any "Euler characteristic" $\chi$ on such a category must be zero, because for any object $P$, we have $$P \oplus \bigoplus_{i=1}^\infty P \simeq \bigoplus_{i = 1}^\infty P$$ which implies that $\chi(P) + \chi(Q) = \chi(Q)$ for $Q = \bigoplus_{i=1}^\infty P$.

Is there any analog for the higher $K_i$ (of, say, an exact category in Quillen's sense)? I don't know any simple way of thinking of the higher $K_i$ (e.g. via Euler characteristics), but it would be interesting if, say, the associated K-theory space somehow had to be contractible.

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We'll give a very general answer to this in the class as well. –  Clark Barwick Feb 22 '12 at 0:43

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up vote 14 down vote accepted

Yes, there is an analogue, see Weibel's book, chapter V, \S 1.9 (http://www.math.rutgers.edu/~weibel/Kbook.html). He calls an exact category $A$ "flasque" if there is a functor $\infty: A \to A$ such that $A \coprod \infty(A) \cong \infty (A)$ and proves that the Quillen K-theory space $K(A)$ of a flasque category $A$ is contractible. Your countable coproduct gives an example. Analogous statements hold more generally for Waldhausen categories and Waldhausen K-theory.

Another result that fits into this context is that if $R$ is a ring, the group of ALL automorphisms of $R^{\infty}$ is acylic (Mac Duff, de la Harpe, "Acyclic groups of automorphisms"). This is proven by an Eilenberg-swindle argument as well.

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Thanks for this answer. –  Akhil Mathew Feb 21 '12 at 21:23

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