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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #7517 by Hao Peng on

Comment #7649 by Stacks Project on