The Stacks project

Should the lemma say instead that the composition is the identity map? Or is the first map $K_0(\mathcal{B}) \rightarrow K_0(D^b_{\mathcal{B}}(\mathcal{A}))$ not the map $[B] \mapsto [B[0]]$?

Unless I have misunderstood something, it also seems to me that the natural map $K_0(\mathcal{B}) \rightarrow K_0(D^b_{\mathcal{B}}(\mathcal{A}))$ given by $[B] \mapsto [B[0]]$ is a group isomorphism, with inverse as described in this lemma.

To see that the composite $K_0(D^b_{\mathcal{B}}(\mathcal{A})) \rightarrow K_0(\mathcal{B}) \rightarrow K_0(D^b_{\mathcal{B}}(\mathcal{A}))$ is the identity map, we can use the distinguished triangle of Remark 13.12.4 and repeatedly take canonical truncations for any bounded complex $X$ in $D^b_{\mathcal{B}}(\mathcal{A})$.

This would generalize Lemma 13.28.2.

Very good indeed. I have made the corresponding changes in this commit.