Lemma 13.28.5. Let $\mathcal{B}$ be a weak Serre subcategory of the abelian category $\mathcal{A}$. There is a canonical isomorphism

$K_0(\mathcal{B}) \longrightarrow K_0(D^ b_\mathcal {B}(\mathcal{A})),\quad [B] \longmapsto [B]$

The inverse sends the class $[X]$ of $X$ to the element $\sum (-1)^ i[H^ i(X)]$.

Proof. We omit the verification that the rule for the inverse gives a well defined map $K_0(D^ b_\mathcal {B}(\mathcal{A})) \to K_0(\mathcal{B})$. It is immediate that the composition $K_0(\mathcal{B}) \to K_0(D^ b_\mathcal {B}(\mathcal{A})) \to K_0(\mathcal{B})$ is the identity. On the other hand, using the distinguished triangles of Remark 13.12.4 and an induction argument the reader may show that the displayed arrow in the statement of the lemma is surjective (details omitted). The lemma follows. $\square$

Comment #7815 by Anonymous on

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 $[B] \mapsto [B]$?

Comment #7833 by Anonymous on

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 $[B] \mapsto [B]$ 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.

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