For an additive category $\mathcal{A}$, how does one show $K_0(\mathcal{A})\cong K_0(\mathcal{K}^b(\mathcal{A}))$?

Question

This is an exercise in §3.13 Beilinson's notes on homological algebra. He doesn't specify but I'm pretty sure $K_0(\mathcal{A})$ is defined as the free group on the isomorphism classes of $\mathcal{A}$ modulo the relations generated by finite (co)products, e.g. $[a\oplus b]=[a]+[b]$, whereas $\mathcal{K}^b(\mathcal{A})$ is the bounded homotopy category with triangulation given by the cofiber sequences and its $K_0$ is the free group on isomorphism (that is, homotopy equivalence) classes of chain complexes modulo the relation that for any exact triangle $A^\bullet\rightarrow B^\bullet\rightarrow C^\bullet\rightarrow A^\bullet[1]$, we have $[A^\bullet]+[C^\bullet]=[B^\bullet]$.

I think the map going $K_0(\mathcal{A})\rightarrow K_0(\mathcal{K}^b(\mathcal{A}))$ should send an object $a$ to the homotopy equivalence class containing its corresponding complex concentrated in degree 0, whereas the inverse will be an "Euler characteristic" map $A_{\bullet}\mapsto\Sigma(-1)^iA_i$. I'm stuck on showing the injectivity of the latter map.

I think the crucial property to use should be the fact that for any two chain morphisms $f,g:A^\bullet\rightarrow B^\bullet$, in $K_0(\mathcal{K}^b(\mathcal{A}))$ we have $[\text{cone}(f)]=[B^\bullet]-[A^\bullet]=[\text{cone}(g)]$, which somehow tells us that the differential of a complex doesn't matter very much in determining its class in the $K$-group. Setting $B^\bullet=A^\bullet$ and $f=0, g=\text{id}$ we can use the contractibility of $\text{cone}(\text{id})$ to show that $[A^\bullet[1]]=-[A^\bullet]$, so if we prove that in $K_0(\mathcal{K}^b(\mathcal{A}))$ every complex is in the same class as its replacement where we've killed the differential, the required result will follow. What I'm not sure about is how to use the relations in $K_0(\mathcal{K}^b(\mathcal{A}))$ to obtain this. Any help would be much appreciated.

Answer 1

Consider the maps \begin{align*} i \colon K_0(\mathscr A) &\to K_0\big(K^{\text{b}}(\mathscr A)\big) & & & \chi \colon K_0\big(K^{\text{b}}(\mathscr A)\big) &\to K_0(\mathscr A)\\ [A] &\mapsto\big [A[0]\big] & & & \big[K^*\big] &\mapsto \sum_i (-1)^i \big[K^i\big]. \end{align*} It is clear that $i$ is well-defined, and for $\chi$ one can use the equivalent definition of the distinguished triangles through termwise split short exact sequences [Tag 014Q]. (To get a termwise split sequence from a mapping cone sequence, use the mapping cylinder. This is explained (poorly) in [Tag 014L].)

Clearly $\chi \circ i = \operatorname{id}$, so it suffices to show that $i \circ \chi = \operatorname{id}$. We prove this by induction on the number $n$ of nonzero terms of $K^*$. If $n \leq 1$, then $K^* = A[i]$ for some $i$, and the result follows since $[A[i]] = (-1)^i [A]$. In general, let $K^*$ be a bounded complex in degrees $[a,b]$ (with $b-a+1 = n$), and consider the stupid truncation $\sigma_{>a}K^*$ [Tag 0118], which sits in a termwise split short exact sequence $$0 \to \sigma_{>a}K^* \to K^* \to K^a[-a] \to 0.$$ This gives $[K^*] = [\sigma_{>a}K^*] + [K^a[-a]]$, so we proceed by induction. $\square$