Quasi-isomorphism from "almost acyclic" complex to its homology

Question

The following is an exercise in the book Representation Theory of Finite Reductive Groups by Cabanes and Enguehard.

Let $\mathcal{A}$ be an abelian category. Let $X$ be a complex of objects of $\mathcal{A}$. Assume $H^i (X ) = 0$ for all $i \neq 0$. Show that $X$ is quasi-isomorphic to $H(X)$. Generalize with an interval $\subseteq \mathbb{Z}$ instead of $\{0\}$, and truncation instead of $H(X)$.

The complex $H(X)$ is the complex with $H^i(X)$ in degree $i$ and zero differentials.

My problem is that I fail to find a natural map $X^0 \rightarrow H^0(X)$ as I believe I am supposed to. Of course, the natural map $X \rightarrow \tau_{\geq 0} X$ to the truncated complex is a quasi-isomorphism under the above conditions and so is $H(X) = \tau_{\leq 0}\tau_{\geq 0} X \rightarrow \tau_{\geq 0} X$. However, being quasi-isomorphic is not an equivalence relation, so this does not suffice. Any help is appreciated.

Answer 1

Consider the complex $\def\ZZ{\mathbb{Z}}\cdots\to0\to\ZZ/4\ZZ\to\ZZ/2\ZZ\to0\to\cdots$, with the nonzero map surjective and the group of order 4 in degree zero. There is exactly one nonzero map $X\to H(X)$ but it doesn't induce an isomorphism in homology.