The Stacks project

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. $\square$