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Let $R$ be a ring and let $\mathcal{C}$ be the category of perfect $R$-complexes. Suppose that $$S=\bigoplus_{i=1}^{\infty}R$$

Let us define $\mathcal{D}$ the smallest thick category generated by $S$.

Clearly $\mathcal{C}$ is a full subcategory of $\mathcal{D}$

The natural embedding $i:\mathcal{C}\rightarrow \mathcal{D} $ induces a morphism in algebraic $K$-theory given by $K(i):K(\mathcal{C})\rightarrow K(\mathcal{D}) $

My question is the following: What can be said about $K_{n}(\mathcal{C})\rightarrow K_{n}(\mathcal{D}) $ for each $n\in \mathbf{N}$, is it an injective/surjective homomorphism?

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    $\begingroup$ K-theory of $\mathcal{D}$ is just zero due to the Eilenberg swindle, or am I missing something? $\endgroup$ Jun 30 at 16:05
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Following Achim's comment, it suffices to show that one can apply the usual Eilenberg swindle argument.

Maybe it's not super clear to the OP that D is closed under countable direct sums of any single object (which is all that one needs for the Eilenberg swindle).

The point is as follows : let $E$ be the subcategory of $D$ on those objects $X$ such that $\bigoplus_{i\in\mathbb N} X \in D$. Then $E$ is a thick subcategory of $D$, and it contains $S$ (because $\mathbb N\times \mathbb N \sim \mathbb N$), thus it contains $D$. In particular $\bigoplus_\mathbb N : D\to D$ is well-defined and allows the Eilenberg swindle argument to go through.

For the sake of self-containment, let me recall how this argument goes : $\bigoplus_\mathbb N \simeq \mathrm{id}_D \oplus \bigoplus_\mathbb N$.

Therefore by additivity, $\mathrm{id}_{K(D)} + K(F) \simeq K(F)$ (where $F = \bigoplus_\mathbb N$), and so $\mathrm{id}_{K(D)} \simeq 0$, in particular $K(D)=0$.

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