Elementwise formula for the connecting map in module categories

Because $c$ is a cycle, $p_{n-1}(d_n^{B}b)=d_n^C(p_{n}b)=d_n^C(c)=0.$ By [L2] and [L3], $d_n^{B}b$ lies in the image of $i_{n-1}$, so there is a unique $a\in A_{n-1}$ with $i_{n-1}(a)=d_n^{B}b$. The definition in [L1] then gives ${\mathrm{\partial }}_n([c])=[a]$.

If $b^{\mathrm{\prime }}$ is another lift of the same cycle $c$, then $b-b^{\mathrm{\prime }}=i_n(u)$ for some $u\in A_n$ by [L2] and [L3]. Hence $i_{n-1}(a-a^{\mathrm{\prime }})=d_n^B(b-b^{\mathrm{\prime }})=d_n^B(i_n(u))=i_{n-1}(d_n^{A}u)\mathrm{.}$ Since $i_{n-1}$ is injective, $a-a^{\mathrm{\prime }}=d_n^{A}u$ is a boundary, so $[a]=[a^{\mathrm{\prime }}]$.

If $c^{\mathrm{\prime }}=c+d_{n+1}^C(v)$ is another cycle representative of $[c]$, choose a lift $v^{\mathrm{\prime }}\in B_{n+1}$ of $v$. Then $b+d_{n+1}^B(v^{\mathrm{\prime }})$ lifts $c^{\mathrm{\prime }}$, and its boundary differs from $d_n^B(b)$ by $d_n^B{d}_{n+1}^B(v^{\mathrm{\prime }})=0.$ Equivalently, the corresponding element of $A_{n-1}$ differs from $a$ by a boundary. So the class from step 1.1 depends only on $[c]$.