Shift signs and exact-functor maps on triangulated K0

The identity triangle $X\stackrel{1_X}{\to }X\to 0\to X[1]$ is distinguished by TR1, so its relation $[X]=[X]+[0]$ holds in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(\mathcal{T})$; subtracting $[X]$ gives $[0]=0$.

Let $F:\mathcal{T}\to {\mathcal{T}}^{\mathrm{\prime }}$ be exact. The assignment $\phi ([X]):=[F(X)]$ is a function on $\mathrm{Iso}(\mathcal{T})$: a functor preserves isomorphisms, so isomorphic objects have isomorphic images. If $X\to Y\to Z\to X[1]$ is distinguished, exactness of $F$ makes $F(X)\to F(Y)\to F(Z)\to F(X)[1]$ distinguished, using the specified shift isomorphism, so $\phi$ satisfies $\phi ([Y])=\phi ([X])+\phi ([Z])$. By the free-group and quotient universal properties of [F6], in the pattern recalled in [F7], $\phi$ factors uniquely through a homomorphism $F_{\ast }:K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}(\mathcal{T})\to K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}({\mathcal{T}}^{\mathrm{\prime }})$ with $F_{\ast }([X])=[F(X)]$.

By TR2 the signed left rotation $X\to 0\to X[1]\stackrel{-1_{X[1]}}{\to }X[1]$ of the identity triangle is distinguished, so $[0]=[X]+[X[1]]$; by step 1.1, $[X[1]]=-[X]$. Applying this to $X[-1]$, and using that $[-1]$ is a quasi-inverse of $[1]$ so that $(X[-1])[1]\cong X$, gives $[X]=-[X[-1]]$ and hence $[X[-1]]=-[X]$.

For the identity functor $1_{\mathcal{T}}$ the class function $[X]\mapsto [1_{\mathcal{T}}(X)]$ is $[X]\mapsto [X]$, so $(1_{\mathcal{T}}{)}_{\ast }=1$. If $F:\mathcal{T}\to {\mathcal{T}}^{\mathrm{\prime }}$ and $G:{\mathcal{T}}^{\mathrm{\prime }}\to {\mathcal{T}}^{\mathrm{\prime }\mathrm{\prime }}$ are exact, then $G\circ F$ is exact: additivity, the composite natural isomorphism ${\xi }_{F(X)}^G\circ G({\xi }_X^F):GF(X[1])\to GF(X)[1]$, and preservation of distinguished triangles all compose. On every generator, $(G\circ F{)}_{\ast }([X])=[(G\circ F)(X)]=[G(F(X))]=G_{\ast }(F_{\ast }([X]))$, so $(G\circ F{)}_{\ast }=G_{\ast }\circ F_{\ast }$.

Let $\eta :F\Rightarrow G$ be a natural isomorphism between exact functors. Each component ${\eta }_X:F(X)\to G(X)$ is an isomorphism by [F5], so $F(X)$ and $G(X)$ have the same isomorphism class in ${\mathcal{T}}^{\mathrm{\prime }}$ and hence $[F(X)]=[G(X)]$ in $K_0^{\mathrm{t}\mathrm{r}\mathrm{i}}({\mathcal{T}}^{\mathrm{\prime }})$. The two induced homomorphisms agree on every generator of the free group and therefore are equal. This is an equality of group homomorphisms only; no coherence for a collection of such natural isomorphisms, and no group-action data, is asserted or obtained.

For $n\ge 0$, induction on $n$ using $[X[n]]=[(X[n-1])[1]]=-[X[n-1]]$ gives $[X[n]]=(-1{)}^n[X]$; the case $n=0$ uses $X[0]\cong X$ and $n=1$ is step 2.1. For $n<0$, apply the nonnegative case to $X[n]$: $[X]=[(X[n])[-n]]=(-1{)}^{-n}[X[n]]$, so $[X[n]]=(-1{)}^n[X]$. Every integer $n$ is covered by the two cases.