A free PID complex decomposes into two-term cycle-boundary pieces

Both $Z_{n}C$ and $B_{n}C$ are submodules of the free module $C_n$, the latter lying in the former since $d_n{d}_{n+1}=0$. Apply the local submodule lemma to get their freeness for every $n$. Thus the second sequence in [F1] is a length-one free presentation of $H_{n}C$.

The surjection $d_n:C_n\twoheadrightarrow B_{n-1}C$ admits a lift of the identity of its free target, hence a section $s_n$. For each $n$ the set of such sections is nonempty; AC selects one for every degree. At $n=0$, take the unique map $s_0:0\to C_0$.

Define $u_n(z,b)=z+s_n(b)$ and $v_n(c)=(c-s_n(d_{n}c),d_{n}c)$. The first component of $v_n(c)$ is a cycle because its differential is $d_{n}c-d_n{s}_n(d_{n}c)=0$. Both maps are linear. Substitution gives $u_n{v}_n(c)=c$ and $v_n{u}_n(z,b)=(z,b)$, since $d_{n}z=0$ and $d_n{s}_n(b)=b$. Thus they are inverse isomorphisms.

Compute $d_n{u}_n(z,b)=b$. Since $d_{n-1}b=0$, its image under $v_{n-1}$ is $(b,0)$. For each $p$ let $E^{(p)}$ have $B_{p}C$ in degree $p+1$, $Z_{p}C$ in degree $p$, and differential the inclusion. In degree $n$, ${⨁}_{p\ge 0}{E}^{(p)}$ is $Z_{n}C\oplus B_{n-1}C$ with precisely the differential just computed. The maps $u_n$ therefore form the claimed chain isomorphism. Only two summands occur in each degree.

Inclusion $\iota$ is a chain map because $d_n$ kills $Z_{n}C$. The map $\rho$ is a chain map to the zero-differential complex $A(C)$ because ${\rho }_{n-1}{d}_n=d_{n-1}{d}_n=0$. Its kernel and image in degree $n$ are $Z_{n}C$ and $B_{n-1}C$, respectively. This proves the canonical short exact sequence, and the selected $s_n$ prove degreewise splitting. Those sections need not be chain maps: $d_n{s}_n(b)=b$ can be nonzero. All formulas hold for the zero complex and for degree zero; replacing $C$ throughout by $D$ proves the stated second application.