Zero homology does not make an object zero in the homotopy category

By [L2], the equality $[1_C]=0$ means precisely that the identity map and the zero map define the same homotopy class. That is equivalent to $1_C$ being null-homotopic, which [L1] says is exactly contractibility.

In the complex $E_{\bullet }$, multiplication by $2$ is injective, reduction modulo $2$ is surjective, and $\ker (<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)=2\mathbb{Z}=\mathrm{im}(2),$ so $E_{\bullet }$ is acyclic. If $E_{\bullet }$ were contractible, then step 1.1 would make $[1_E]=0$, equivalently $1_E$ would be null-homotopic. In degree $0$ that would force a section $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ of the quotient map $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which is impossible. Thus $E_{\bullet }$ is not contractible.

Step 2.1 gives an acyclic complex that is not contractible, so by step 1.1 its identity class is not zero in the homotopy category. Therefore vanishing homology alone does not force an object to be zero there. This does not contradict [L4], which gives only the forward implication contractible $\Rightarrow$ acyclic.