The Stacks project

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

\[ c(X) = \sum (-1)^ i[H^ i(X)] \in K_0(\mathcal{A}) \]

Given a distinguished triangle $X \to Y \to Z$ the long exact sequence of cohomology ( 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)

\[ \sigma _{\geq n}X^\bullet \to \sigma _{\geq n - 1}X^\bullet \to X^{n - 1}[-n + 1] \to (\sigma _{\geq n}X^\bullet )[1] \]

and induction show that

\[ [X^\bullet ] = \sum (-1)^ i[X^ i[0]] \]

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. $\square$

Comments (2)

Comment #7517 by Hao Peng on

I think in the proof we should use the usual truncation instead of the stupid truncation.

Comment #7649 by on

@#7517: I do not agree. @everyone: if you think there is an error in a proof, then please point out exactly what the error is.

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