G0 of an abelian category equals triangle K0 of its bounded derived category

The assignment $A\mapsto S^0(A)$, functorial by degreewise application, sends isomorphic objects of $\mathcal{C}$ to isomorphic degree-zero stalk complexes, so $[A]\mapsto [S^0(A)]$ is a well-defined class function on $\mathrm{Iso}(\mathcal{C})$ with values in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$: the stalk complex has cohomology supported in degree $0$ by [F7], hence lies in $D^b(\mathcal{C})$ by [F5]. A short exact sequence $0\to A\to B\to C\to 0$ of $\mathcal{C}$, viewed in degree zero, is degreewise short exact as a sequence of complexes (in every other degree it reads $0\to 0\to 0$ with zero differentials), so [F6] gives a distinguished triangle $S^0(A)\to S^0(B)\to S^0(C)\to S^0(A)[1]$ in $D(\mathcal{C})$ whose objects all have bounded cohomology and therefore lie in $D^b(\mathcal{C})$ by [F5]; its relation $[S^0(B)]=[S^0(A)]+[S^0(C)]$ holds in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$ by [F2]. Thus the class function is additive on short exact sequences, and [F1] gives a unique homomorphism ${\iota }_{\ast }:G_0(\mathcal{C})\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$ with ${\iota }_{\ast }([A])=[S^0(A)]$.

For $X$ in $D^b(\mathcal{C})$ the cohomology objects $H^n(X)$ are defined for all $n$ by [F4] and vanish outside a finite interval by [F5]; set $\chi (X):={\sum }_n(-1{)}^n[H^n(X)]\in G_0(\mathcal{C})$, a finite alternating sum. If $X\cong X^{\mathrm{\prime }}$ in $D^b(\mathcal{C})$, then $H^n(X)\cong H^n(X^{\mathrm{\prime }})$ for every $n$ because the cohomology functors factor through the derived category [F4], so $\chi$ is a class function on $\mathrm{Iso}(D^b(\mathcal{C}))$. The zero object has $\chi (0)=0$, since its cohomology vanishes in every degree.

The class function $\chi$ of step 1.2 is additive on distinguished triangles. Let $X\to Y\to Z\to X[1]$ be a distinguished triangle of $D^b(\mathcal{C})$; its image under the exact inclusion is distinguished in $D(\mathcal{C})$ by [F5], so [F4] gives the long exact cohomology sequence $\cdots \to H^n(X)\to H^n(Y)\to H^n(Z)\to H^{n+1}(X)\to \cdots$. Choose integers $a\le b$ with $H^n(X)=H^n(Y)=H^n(Z)=0$ for $n<a$ and $n>b$; such bounds exist since all three objects lie in the essential image described in [F5], and the sequence vanishes outside $[a,b]$. For each $n$ put $J^n:=\ker (H^n(X)\to H^n(Y))$, $K^n:=\mathrm{im}(H^n(X)\to H^n(Y))$ and $I^n:=\mathrm{im}(H^n(Y)\to H^n(Z))$, objects of $\mathcal{C}$ that are subobjects or quotients of $H^n(X),H^n(Y),H^n(Z)$ and hence vanish for $n<a$ and $n>b$. Exactness at $H^n(X)$, $H^n(Y)$ and $H^n(Z)$, together with $\mathrm{im}(H^n(Z)\to H^{n+1}(X))=\ker (H^{n+1}(X)\to H^{n+1}(Y))=J^{n+1}$, gives the short exact sequences $0\to J^n\to H^n(X)\to K^n\to 0$, $0\to K^n\to H^n(Y)\to I^n\to 0$ and $0\to I^n\to H^n(Z)\to J^{n+1}\to 0$ in $\mathcal{C}$, with $J^{b+1}\subseteq H^{b+1}(X)=0$. Multiplying the three resulting $G_0$-relations by $(-1{)}^n$ and summing over the finitely many nonzero degrees gives $\chi (Y)=\chi (X)+\chi (Z)$: the $J$-terms telescope, since ${\sum }_n(-1{)}^n[J^{n+1}]=-{\sum }_n(-1{)}^n[J^n]$, while the $K$- and $I$-terms cancel between $[H^n(X)]+[H^n(Z)]$ and $[H^n(Y)]$.

Since $\chi$ is a class function additive on distinguished triangles by step 2.1, extending it over the free abelian group on $\mathrm{Iso}(D^b(\mathcal{C}))$ and applying the quotient universal property, exactly as the functor-induced class functions of [F2] are factored, produces a unique homomorphism $\bar{\chi }:K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))\to G_0(\mathcal{C})$ with $\bar{\chi }([X])=\chi (X)$.

The composite $\bar{\chi }\circ {\iota }_{\ast }$ is the identity of $G_0(\mathcal{C})$: for an object $A$ of $\mathcal{C}$, $\bar{\chi }({\iota }_{\ast }([A]))=\chi (S^0(A))={\sum }_n(-1{)}^n[H^n(S^0(A))]=[A]$ by [F7], so the two homomorphisms agree on every generator of $G_0(\mathcal{C})$ [F1], where ${\iota }_{\ast }$ is the homomorphism of step 1.1 and $\bar{\chi }$ is that of step 3.1.

The composite ${\iota }_{\ast }\circ \bar{\chi }$ is the identity of $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$. Let $X$ have $H^n(X)=0$ for $n<a$ and $n>b$. The canonical truncations are objects of $D^b(\mathcal{C})$ and the triangles ${\tau }^{\le n}X\to {\tau }^{\le n+1}X\to H^{n+1}(X)[-n-1]\to ({\tau }^{\le n}X)[1]$ of [F6] are distinguished in $D^b(\mathcal{C})$ by [F5]. The complex ${\tau }^{\le a-1}X$ has zero cohomology in every degree by [F6], so the zero map ${\tau }^{\le a-1}X\to 0$ is a quasi-isomorphism and $[{\tau }^{\le a-1}X]=0$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$. For each $n=a-1,\dots ,b-1$ the triangle relation and $[H^{n+1}(X)[-n-1]]=(-1{)}^{n+1}[H^{n+1}(X)[0]]$ from [F2] give $[{\tau }^{\le n+1}X]=[{\tau }^{\le n}X]+(-1{)}^{n+1}[H^{n+1}(X)[0]]$, and summing these relations telescopes to $[{\tau }^{\le b}X]={\sum }_{n=a}^b(-1{)}^n[H^n(X)[0]]$. The truncation map ${\tau }^{\le b}X\to X$ is a quasi-isomorphism by the cohomology formula of [F6], so $[{\tau }^{\le b}X]=[X]$ and $[X]={\sum }_{n=a}^b(-1{)}^n[H^n(X)[0]]={\sum }_{n=a}^b(-1{)}^n{\iota }_{\ast }([H^n(X)])={\iota }_{\ast }(\bar{\chi }([X]))$, using the identification ${\iota }_{\ast }([H^n(X)])=[S^0(H^n(X))]=[H^n(X)[0]]$ of step 1.1. The classes $[X]$ generate $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$ [F2], so ${\iota }_{\ast }\circ \bar{\chi }$ is the identity.

Steps 4.1 and 4.2 exhibit $\bar{\chi }$ as a two-sided inverse of ${\iota }_{\ast }$, so degree-zero inclusion induces the asserted isomorphism $G_0(\mathcal{C})\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(D^b(\mathcal{C}))$ whose inverse sends $[X]$ to ${\sum }_n(-1{)}^n[H^n(X)]$. The argument used only the free abelian group presentations, their universal properties, the long exact cohomology sequence and the finite canonical-truncation induction: no enough-projectives or enough-injectives hypothesis, no Noetherianity and no finite global dimension was used, and no choice principle occurs.