Let $A^\bullet$ be the complex
$$\ldots \rightarrow A^{-2} \rightarrow A^{-1} \rightarrow A^0 \rightarrow A^1 \rightarrow \ldots,$$ where the $A^i$'s are objects of an abelian category $\mathcal{A}$. We have differential maps $d^i:A^i \rightarrow A^{i+1}$, for all $i$.
I briefly recall the truncation functors:
We have an endo-functor $\tau_{\leq n}$ on $\mathbf{C}(\mathcal{A})$, which sends $A^\bullet$ to the complex $$\tau_{\leq n}(A^\bullet) = \ldots \rightarrow A^{n-2} \rightarrow A^{n-1} \rightarrow \mathrm{ker}(d^n) \rightarrow 0 \rightarrow \ldots.$$
This sort of cuts the complex at the term of degree $n$, and it maintains the $i$-th cohomology for $i \leq n$, while the others are $0$.
Dually, we also have the truncation functor $\tau_{\geq n}$ which sends $A^\bullet$ to $$\tau_{\geq n}(A^\bullet) = \ldots \rightarrow 0 \rightarrow \mathrm{coker}(d^{n-1}) \rightarrow A^{n+1} \rightarrow A^{n+2} \rightarrow \ldots.$$
We have an exact sequence of complexes $$0 \rightarrow \tau_{\leq n-1}(A^\bullet) \rightarrow A^\bullet \rightarrow \tau_{\geq n}(A^\bullet) \rightarrow 0.$$
This is where I encounter a problem: the $(n-1)$-th row (i.e., the exact sequence of degree $n-1$ terms) would read as $$0 \rightarrow \mathrm{ker}(d^{n-1}) \rightarrow A^{n-1} \rightarrow 0 \rightarrow 0$$ which is exact if and only if $d^{n-1}$ is the zero map. Similarly, the $n$-th row is $$0 \rightarrow 0 \rightarrow A^n \rightarrow \mathrm{coker}(d^{n-1}) \rightarrow 0$$ which again is exact if and only if $d^{n-1}$ is zero.
What is going on here?
Best Answer
There are two truncation functors :
$$\tau_{\leq n}(A^\bullet) = ...\to A^{n-1}\to\ker d^n\to 0\to 0\to...$$ $$\tau'_{\leq n}(A^\bullet) = ...\to A^{n-1}\to A^n\to\operatorname{im}d^n\to 0\to...$$
and the canonical inclusion $\tau_{\leq n}(A^\bullet)\to\tau'_{\leq n}(A^\bullet)$ is a natural quasi-isomorphism (hence an isomorphism in the derived category). This is why they are often denoted by the same symbol.
Dually you have $$\tau_{\geq n}(A^\bullet)=...\to 0\to 0\to\operatorname{coker}d^{n-1}\to A^{n+1}\to ...$$ $$\tau'_{\geq n}(A^\bullet)=...\to 0\to\operatorname{im}d^{n-1}\to A^n\to A^{n+1}\to ...$$ and similarily a natural quasi-isomorphism $\tau_{\geq n}(A^\bullet)\to \tau'_{\geq n}(A^\bullet)$.
The short exact sequence is often written as a distinguished triangle in the derived category. But it can also be seen in the category of complexes, but then you should take $$0\to \tau_{\leq n-1}(A^\bullet)\to A^\bullet \to \tau'_{\geq n}(A^\bullet)\to 0$$ or $$0\to \tau'_{\leq n-1}(A^\bullet)\to A^\bullet \to \tau_{\geq n}(A^\bullet)\to 0$$