FALSE: an acyclic mapping cone is contractible

The displayed source complex is acyclic: multiplication by $2$ is injective, reduction modulo $2$ is surjective, and its kernel is $2\mathbb{Z}$, the image of the first map. If it were contractible, [L1] would make its identity null-homotopic; in degree $0$ that would supply a section $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ of the quotient map $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which is impossible. Thus the source complex is acyclic and noncontractible.

By [L2], the displayed mapping cone is the shift of the source complex. Because the zero map from an acyclic complex to the zero complex induces isomorphisms on all homology groups, [L4] makes that map a quasi-isomorphism; then [L3] makes its shift a quasi-isomorphism too. Hence the cone is acyclic. If it were contractible, applying [L3] with shift $[-1]$ would make the source complex contractible, contradicting step 1.1. Therefore the displayed mapping cone is acyclic but not contractible, directly contradicting [A1]. This is why the cone criteria for quasi-isomorphism and homotopy equivalence are different.