Lemma 13.28.2. Let \mathcal{A} be an abelian category. Then there is a canonical identification K_0(D^ b(\mathcal{A})) = K_0(\mathcal{A}) of zeroth K-groups.
Proof. Given an object A of \mathcal{A} denote A[0] the object A viewed as a complex sitting in degree 0. If 0 \to A \to A' \to A'' \to 0 is a short exact sequence, then we get a distinguished triangle A[0] \to A'[0] \to A''[0] \to A[1], see Section 13.12. This shows that we obtain a map K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A})) by sending [A] to [A[0]] with apologies for the horrendous notation.
On the other hand, given an object X of D^ b(\mathcal{A}) we can consider the element
Given a distinguished triangle X \to Y \to Z the long exact sequence of cohomology (13.11.1.1) and the relations in K_0(\mathcal{A}) show that c(Y) = c(X) + c(Z). Thus c factors through a map c : K_0(D^ b(\mathcal{A})) \to K_0(\mathcal{A}).
We want to show that the two maps above are mutually inverse. It is clear that the composition K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A})) \to K_0(\mathcal{A}) is the identity. Suppose that X^\bullet is a bounded complex of \mathcal{A}. The existence of the distinguished triangles of “stupid truncations” (see Homology, Section 12.15)
and induction show that
in K_0(D^ b(\mathcal{A})) (with again apologies for the notation). It follows that the composition K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A})) is surjective which finishes the proof. \square
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Comment #7517 by Hao Peng on
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