My ultimate goal is to prove that given a heart $\mathcal{C}$ of a bounded $t$-structure $(\mathcal{T}^{\leq 0}, \mathcal{T}^{\geq 0})$ in a triangulated category $\mathcal{T}$, that the Grothendieck groups are isomorphic $K\mathcal{T}\cong K\mathcal{C}$, where the former is defined in the sense of triangulated categories, and the latter in terms of Abelian categories.
For reference, this is Proposition A.9.5 of Achar's book on Perverse Sheaves and Representation theory, the exercise being left to the reader.
Now, I know that the zeroth $t$-chomology functor ${}^tH^0$ is a cohomological functor, and that for all pairs of arrows $X\xrightarrow{f}Y\xrightarrow{g}Z$ in $\mathcal{C}$, the sequence $0\to X\xrightarrow{f}Y\xrightarrow{g}Z\to 0$ is exact iff there's a morphism $h$ making
$X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}$ into a distinguished triangle. Thus, inspired by the notion of Euler characteristic, I hoped that for $X\in \mathcal{T}$, letting $[-]$ denote the isomorphism class of $-$, that the map
$[X]\mapsto \sum_{\mathbb{Z}}(-1)^n [{}^tH^n(X)]$ would serve as the inverse to the map induced by the inclusion $\mathcal{C}\to \mathcal{T}$. However it's here I'm stuck….
Of course, I thought to just consider the case of the bounded derived category $D^b(\mathcal{A})$ of an Abelian category $\mathcal{A}$, and the heart of the natural $t$-structure first, and then let that inspire the approach. (That is, complexes up to quasi-iso and the heart of complexes concentrated Though even here, I'm stuck for what to do. While it makes "spiritual" sense to me that for a complex $A$ we should find $\sum_{\mathbb{Z}}(-1)^n [H^n(A)]=[A]$ (after including back into $KD^b(\mathcal{A})$) I'm not sure how to prove this.
Achar says the proof of that A.9.5 is similar to the very simple proof of a theorem he offers shortly before, but I just don't see it…
Best Answer
As you've already argued why the canonical map $K\mathcal{C} \to K\mathcal{T}$ is injective, it remains to argue surjectivity. In other words, given an object $X$ of $\mathcal{T}$, we have to argue that $[X] \in K\mathcal{T}$ lies in the image of $K\mathcal{C}$.
First assume that $X \in \mathcal{T}_{=n}$ is concentrated in a single degree $n$. Using the exact sequence \begin{aligned} X \to 0 \to X[1], \end{aligned} one sees that $[X] = -[X[1]]$ in $K\mathcal{T}$, showing that $[X]$ is in the image of $K\mathcal{C}$ if and only if $[X[1]]$ is. By induction, this shows the claim when $X \in \mathcal{T}_{=n}$.
Now consider an arbitrary object $X$ of $\mathcal{T}$. Since $\mathcal{T}$ is bounded, it lives in a range $[n,m]$ for some integers $m \geq n$. Fixing $n$, we will prove by induction on $m$ that $[X] \in K\mathcal{T}$ lies in the image of $K\mathcal{C}$. The case $m = n$ was treated before, so assume $m > n$ and assume we've proved the claim for $m - 1$. Now pick a distinguished triangle \begin{aligned} X_{\geq m} \to X \to X_{< m}, \end{aligned} with $X_{\geq m}\in \mathcal{T}_{\geq m}$ and $X_{< m} \in \mathcal{T}_{< m}$. Then the statement holds for $X_{\geq m} \in \mathcal{T}_{=m}$ and $X_{<m} \in \mathcal{T}_{[n,m-1]}$ by the previous step and the induction hypothesis, respectively. Since we have $[X] = [X_{\geq m}] + [X_{<m}]$ in $K\mathcal{T}$, this finishes the proof.