The isolated differential 2 on the integers cannot be cancelled

The entry $2$ has no inverse in the category of abelian groups. If $u:\mathbb{Z}\to \mathbb{Z}$ satisfied $2u=1$ or $u\cdot 2=1$, then evaluating at $1$ gives $2u(1)=1$ with $u(1)\in \mathbb{Z}$, which is impossible because $1$ is odd; equivalently $\mathbb{Z}$ has no element $m$ with $2m=1$. Since $2$ is not invertible, it is not a pivot in the sense of [L1], and the Schur complement $a-b{\phi }^{-1}c$ of the block $(\phi )=(2)$ is not defined on its own.

The complex nonetheless has nonzero homology. Reindexing by $C_m:=X^{-m}$ gives the chain complex $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ concentrated in degrees $-n$ and $-n-1$. There $Z_{-n-1}=\ker (d_{-n-1}=0)=\mathbb{Z}$ and $B_{-n-1}=\mathrm{im}(d_{-n})=\mathrm{im}(2)=2\mathbb{Z}$, so the boundary-to-cycle map is the inclusion $2\mathbb{Z}\hookrightarrow \mathbb{Z}$ and, by [L4], $H_{-n-1}\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}\mathrm{\ne }0$; this is the homology in cochain degree $n+1$ of $X^{\bullet }$.

The complex is not contractible. If it were, [L3] would give a homomorphism $h^{n+1}:\mathbb{Z}\to \mathbb{Z}$ with $1_{X^n}=d^{n-1}{h}^n+h^{n+1}{d}^n=0+h^{n+1}\cdot 2$ in degree $n$, because $d^{n-1}=0$ and $d^n=\cdot 2$; evaluating at $1$ would produce an integer $h^{n+1}(1)$ with $2h^{n+1}(1)=1$, which is impossible by step 1.1. Hence $X^{\bullet }$ is not homotopy equivalent to the zero complex, and deleting both terms would not be a homotopy equivalence.

Consequently the pivot hypothesis of [L1] and [L2] is genuinely needed: the deleted terms carry the nonzero homology object $H_{-n-1}\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$ of step 1.2, which the zero complex does not have, and no two-sided inverse of $2$ exists in $\mathbb{Z}$. The claim refuted is therefore false; the theorem's conclusion is not available here because its hypothesis fails. ∎