[proofplan] We verify that the standard construction of the connecting map is independent of every choice. First, a cycle in $C_n$ and a lift in $B_n$ produce a uniquely determined cycle in $A_{n-1}$ by exactness and the chain map identities. Then we show that changing the lift changes this cycle by a boundary, and changing the representative of the homology class also changes it by a boundary. Finally, choosing lifts additively proves that the resulting map on homology is an $R$-module homomorphism. [/proofplan] [step:Lift a cycle in $C_n$ to a cycle in $A_{n-1}$] Fix $n \in \mathbb Z$. Let $c \in C_n$ be a cycle, so $d^C_n c=0$. Since $p_n:B_n \to C_n$ is surjective, choose $b \in B_n$ with $p_n(b)=c$. Because $p$ is a chain map, \begin{align*} p_{n-1}(d^B_n b) = d^C_n(p_n b) = d^C_n c = 0. \end{align*} Thus $d^B_n b \in \ker p_{n-1}$. Exactness of \begin{align*} 0 \longrightarrow A_{n-1} \xrightarrow{i_{n-1}} B_{n-1} \xrightarrow{p_{n-1}} C_{n-1} \longrightarrow 0 \end{align*} gives $\ker p_{n-1}=\operatorname{im} i_{n-1}$, so there exists $a \in A_{n-1}$ such that \begin{align*} i_{n-1}(a)=d^B_n b. \end{align*} This element $a$ is unique because $i_{n-1}$ is injective. We now prove that $a$ is a cycle. Since $i$ is a chain map, \begin{align*} i_{n-2}(d^A_{n-1}a) = d^B_{n-1}(i_{n-1}a) = d^B_{n-1}(d^B_n b) = 0, \end{align*} because $B_*$ is a chain complex. Since $i_{n-2}$ is injective, $d^A_{n-1}a=0$. Hence $a \in \ker d^A_{n-1}$, so $[a] \in H_{n-1}(A_*)$ is defined. [/step] [step:Show that changing the lift changes $a$ by a boundary] Let $b,b' \in B_n$ be two lifts of the same cycle $c \in C_n$, so $p_n(b)=p_n(b')=c$. Let $a,a' \in A_{n-1}$ be the elements determined by \begin{align*} i_{n-1}(a)=d^B_n b, \qquad i_{n-1}(a')=d^B_n b'. \end{align*} Then \begin{align*} p_n(b'-b)=p_n(b')-p_n(b)=c-c=0. \end{align*} Exactness in degree $n$ gives $\ker p_n=\operatorname{im} i_n$, so there exists $x \in A_n$ such that \begin{align*} i_n(x)=b'-b. \end{align*} Using that $i$ is a chain map, we compute \begin{align*} i_{n-1}(a'-a) &= i_{n-1}(a')-i_{n-1}(a) \\ &= d^B_n b' - d^B_n b \\ &= d^B_n(b'-b) \\ &= d^B_n(i_n x) \\ &= i_{n-1}(d^A_n x). \end{align*} Since $i_{n-1}$ is injective, \begin{align*} a'-a=d^A_n x. \end{align*} Thus $a$ and $a'$ differ by a boundary in $A_{n-1}$, so $[a']=[a]$ in $H_{n-1}(A_*)$. [guided] The only possible ambiguity at this stage is the choice of the lift $b$. If $b$ and $b'$ both lift the same element $c$, then their difference lies over $0$: \begin{align*} p_n(b'-b)=p_n(b')-p_n(b)=c-c=0. \end{align*} Exactness says precisely that every element of $\ker p_n$ comes from $A_n$. Therefore there is an element $x \in A_n$ with \begin{align*} i_n(x)=b'-b. \end{align*} Now compare the two elements $a,a' \in A_{n-1}$ produced from the two lifts. Their images under $i_{n-1}$ are the boundaries of the lifts: \begin{align*} i_{n-1}(a)=d^B_n b, \qquad i_{n-1}(a')=d^B_n b'. \end{align*} Subtracting and using the chain map identity $d^B_n i_n=i_{n-1}d^A_n$ gives \begin{align*} i_{n-1}(a'-a) &= d^B_n b' - d^B_n b \\ &= d^B_n(b'-b) \\ &= d^B_n(i_n x) \\ &= i_{n-1}(d^A_n x). \end{align*} Since $i_{n-1}$ is injective, the equality after applying $i_{n-1}$ already holds in $A_{n-1}$: \begin{align*} a'-a=d^A_n x. \end{align*} Thus the two cycles differ by an $A$-boundary, so they define the same class in $H_{n-1}(A_*)$. [/guided] [/step] [step:Show that changing the cycle representative changes $a$ by a boundary] Let $c,c' \in C_n$ be homologous cycles. Then there exists $y \in C_{n+1}$ such that \begin{align*} c'=c+d^C_{n+1}y. \end{align*} Choose $b \in B_n$ with $p_n(b)=c$. Since $p_{n+1}:B_{n+1}\to C_{n+1}$ is surjective, choose $\widetilde y \in B_{n+1}$ with \begin{align*} p_{n+1}(\widetilde y)=y. \end{align*} Define \begin{align*} b' := b+d^B_{n+1}\widetilde y \in B_n. \end{align*} Then, using that $p$ is a chain map, \begin{align*} p_n(b') &= p_n(b)+p_n(d^B_{n+1}\widetilde y) \\ &= c+d^C_{n+1}(p_{n+1}\widetilde y) \\ &= c+d^C_{n+1}y \\ &= c'. \end{align*} Let $a,a' \in A_{n-1}$ be determined by \begin{align*} i_{n-1}(a)=d^B_n b, \qquad i_{n-1}(a')=d^B_n b'. \end{align*} Then \begin{align*} i_{n-1}(a') = d^B_n b' = d^B_n b+d^B_n d^B_{n+1}\widetilde y = d^B_n b = i_{n-1}(a). \end{align*} Since $i_{n-1}$ is injective, $a'=a$. Therefore the construction gives the same class for $c$ and for the homologous representative $c'$. For an arbitrary lift of $c'$, the preceding step shows that the resulting class is the same as the class obtained from the particular lift $b'=b+d^B_{n+1}\widetilde y$. Hence the homology class $[a] \in H_{n-1}(A_*)$ depends only on $[c] \in H_n(C_*)$. [guided] Now we check the second kind of ambiguity: the representative of the homology class in $C_*$. Suppose $c$ and $c'$ represent the same class in $H_n(C_*)$. By definition, their difference is a boundary, so there is an element $y \in C_{n+1}$ such that \begin{align*} c'=c+d^C_{n+1}y. \end{align*} Because $p_{n+1}:B_{n+1}\to C_{n+1}$ is surjective, we may lift $y$ to an element $\widetilde y \in B_{n+1}$ satisfying \begin{align*} p_{n+1}(\widetilde y)=y. \end{align*} Choose a lift $b \in B_n$ of $c$, and set \begin{align*} b' := b+d^B_{n+1}\widetilde y. \end{align*} This is designed so that $b'$ lifts $c'$. Indeed, since $p$ is a chain map, \begin{align*} p_n(b') &= p_n(b)+p_n(d^B_{n+1}\widetilde y) \\ &= c+d^C_{n+1}(p_{n+1}\widetilde y) \\ &= c+d^C_{n+1}y \\ &= c'. \end{align*} Let $a$ be the element produced from $b$, and let $a'$ be the element produced from this particular lift $b'$: \begin{align*} i_{n-1}(a)=d^B_n b, \qquad i_{n-1}(a')=d^B_n b'. \end{align*} Then the extra term in $b'$ disappears after applying $d^B_n$, because $B_*$ is a chain complex: \begin{align*} i_{n-1}(a') = d^B_n b' = d^B_n b+d^B_n d^B_{n+1}\widetilde y = d^B_n b = i_{n-1}(a). \end{align*} Injectivity of $i_{n-1}$ gives $a'=a$. This proves independence for one carefully chosen lift of $c'$. If another lift of $c'$ is chosen, the previous step applies and shows that it changes the resulting element of $A_{n-1}$ only by an $A$-boundary. Therefore the class in $H_{n-1}(A_*)$ depends only on the homology class $[c]$. [/guided] [/step] [step:Prove that the connecting map is $R$-linear] The preceding steps define a function \begin{align*} \partial_n: H_n(C_*) &\longrightarrow H_{n-1}(A_*) \\ [c] &\longmapsto [a]. \end{align*} We prove that it is an $R$-module homomorphism. Let $[c_1],[c_2]\in H_n(C_*)$, and choose cycle representatives $c_1,c_2 \in C_n$. Choose lifts $b_1,b_2 \in B_n$ with \begin{align*} p_n(b_1)=c_1, \qquad p_n(b_2)=c_2. \end{align*} Let $a_1,a_2 \in A_{n-1}$ be determined by \begin{align*} i_{n-1}(a_1)=d^B_n b_1, \qquad i_{n-1}(a_2)=d^B_n b_2. \end{align*} Then $b_1+b_2$ is a lift of $c_1+c_2$, and \begin{align*} i_{n-1}(a_1+a_2) = i_{n-1}(a_1)+i_{n-1}(a_2) = d^B_n b_1+d^B_n b_2 = d^B_n(b_1+b_2). \end{align*} Hence \begin{align*} \partial_n([c_1]+[c_2]) = [a_1+a_2] = [a_1]+[a_2] = \partial_n([c_1])+\partial_n([c_2]). \end{align*} Let $r \in R$ and $[c]\in H_n(C_*)$. Choose a cycle representative $c \in C_n$, a lift $b \in B_n$ with $p_n(b)=c$, and $a \in A_{n-1}$ with $i_{n-1}(a)=d^B_n b$. Then $rb$ is a lift of $rc$, and, since the differentials and $i_{n-1}$ are $R$-linear, \begin{align*} i_{n-1}(ra) = r\,i_{n-1}(a) = r\,d^B_n b = d^B_n(rb). \end{align*} Therefore \begin{align*} \partial_n(r[c]) = [ra] = r[a] = r\,\partial_n([c]). \end{align*} Thus $\partial_n$ is additive and homogeneous over $R$, so it is an $R$-module homomorphism. [/step]