[proofplan] We compute both composites on an arbitrary homology class $[c] \in H_n(C_*)$. Choose a lift $b \in B_n$ of the cycle representative $c$, and let $a \in A_{n-1}$ be the element whose image under $i_{n-1}$ is $d_B b$; by definition, $\partial_n([c]) = [a]$. The commutativity of the morphism of short exact sequences shows that $f_{B,n}(b)$ is a lift of $f_{C,n}(c)$ and that its differential is $i'_{n-1}(f_{A,n-1}(a))$. Therefore the primed connecting homomorphism sends $H_n(f_C)([c])$ to $[f_{A,n-1}(a)]$, which is also $H_{n-1}(f_A)(\partial_n([c]))$. [/proofplan]

[step:Choose a cycle representative and lift it through the short exact sequence]Fix $n \in \mathbb Z$ and let $[c] \in H_n(C_*)$, where $c \in C_n$ is a cycle, so $d_C c = 0$. Since $p_n: B_n \to C_n$ is surjective by exactness, choose $b \in B_n$ such that \begin{align*} p_n(b) = c. \end{align*} Because $p: B_* \to C_*$ is a chain map, \begin{align*} p_{n-1}(d_B b) = d_C(p_n(b)) = d_C c = 0. \end{align*} Thus $d_B b \in \ker p_{n-1}$. Exactness at $B_{n-1}$ 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 b. \end{align*} Since $i_{n-1}$ is injective, this element $a$ is uniquely determined by $b$.[/step]

[guided]Fix an integer $n$ and a homology class $[c] \in H_n(C_*)$. By definition of homology, we may choose a representative cycle $c \in C_n$ satisfying \begin{align*} d_C c = 0. \end{align*} The short exactness of \begin{align*} 0 \longrightarrow A_n \xrightarrow{i_n} B_n \xrightarrow{p_n} C_n \longrightarrow 0 \end{align*} implies that $p_n$ is surjective. Therefore there is an element $b \in B_n$ with \begin{align*} p_n(b) = c. \end{align*} The connecting homomorphism is built by differentiating such a lift. We compute where $d_B b$ lands under $p_{n-1}$. Since $p$ is a chain map, it commutes with differentials: \begin{align*} p_{n-1}(d_B b) = d_C(p_n(b)) = d_C c = 0. \end{align*} Hence $d_B b \in \ker p_{n-1}$. 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*} says precisely that $\ker p_{n-1} = \operatorname{im} i_{n-1}$. Therefore there exists $a \in A_{n-1}$ such that \begin{align*} i_{n-1}(a) = d_B b. \end{align*} The same exactness also gives injectivity of $i_{n-1}$, so this $a$ is uniquely determined once the lift $b$ has been chosen.[/guided]

[step:Identify the first composite using the definition of the connecting homomorphism] By the definition of the connecting homomorphism for the short exact sequence $0 \to A_* \to B_* \to C_* \to 0$, \begin{align*} \partial_n([c]) = [a] \in H_{n-1}(A_*). \end{align*} Since $f_A: A_* \to A'_*$ is a chain map, it induces \begin{align*} H_{n-1}(f_A): H_{n-1}(A_*) \to H_{n-1}(A'_*), \end{align*} and therefore \begin{align*} \bigl(H_{n-1}(f_A) \circ \partial_n\bigr)([c]) = H_{n-1}(f_A)([a]) = [f_{A,n-1}(a)]. \end{align*} [/step]

[step:Show that applying $f_B$ gives a valid lift in the primed sequence]The commutativity relation $p'_n \circ f_{B,n} = f_{C,n} \circ p_n$ gives \begin{align*} p'_n(f_{B,n}(b)) = f_{C,n}(p_n(b)) = f_{C,n}(c). \end{align*} Thus $f_{B,n}(b) \in B'_n$ is a lift of $f_{C,n}(c) \in C'_n$ through $p'_n$. Since $f_C$ is a chain map and $c$ is a cycle, \begin{align*} d_{C'}(f_{C,n}(c)) = f_{C,n-1}(d_C c) = f_{C,n-1}(0) = 0. \end{align*} Hence $f_{C,n}(c)$ is a cycle representing $H_n(f_C)([c]) \in H_n(C'_*)$.[/step]

[guided]To compute the other composite, we must apply the primed connecting homomorphism to the class represented by $f_{C,n}(c)$. For this, we need a lift of $f_{C,n}(c)$ in $B'_n$. The morphism of short exact sequences gives the commutative square involving $B_n$, $C_n$, $B'_n$, and $C'_n$: \begin{align*} p'_n \circ f_{B,n} = f_{C,n} \circ p_n. \end{align*} Applying this identity to the chosen lift $b \in B_n$ yields \begin{align*} p'_n(f_{B,n}(b)) = f_{C,n}(p_n(b)) = f_{C,n}(c). \end{align*} Therefore $f_{B,n}(b)$ is exactly the lift in $B'_n$ needed to compute $\partial'_n$. We also verify that $f_{C,n}(c)$ is a cycle. Since $f_C$ is a chain map, \begin{align*} d_{C'}(f_{C,n}(c)) = f_{C,n-1}(d_C c). \end{align*} Because $c$ is a cycle, $d_C c = 0$, and since $f_{C,n-1}$ is $R$-linear, \begin{align*} f_{C,n-1}(d_C c) = f_{C,n-1}(0) = 0. \end{align*} Thus $f_{C,n}(c)$ represents the homology class $H_n(f_C)([c])$ in $H_n(C'_*)$.[/guided]

[step:Compute the primed connecting homomorphism on the transported class]Because $f_B$ is a chain map, \begin{align*} d_{B'}(f_{B,n}(b)) = f_{B,n-1}(d_B b). \end{align*} Using $d_B b = i_{n-1}(a)$ and the commutativity relation $f_{B,n-1} \circ i_{n-1} = i'_{n-1} \circ f_{A,n-1}$, we obtain \begin{align*} d_{B'}(f_{B,n}(b)) = f_{B,n-1}(i_{n-1}(a)) = i'_{n-1}(f_{A,n-1}(a)). \end{align*} By the definition of the connecting homomorphism for the primed short exact sequence, \begin{align*} \partial'_n([f_{C,n}(c)]) = [f_{A,n-1}(a)] \in H_{n-1}(A'_*). \end{align*} Since $H_n(f_C)([c]) = [f_{C,n}(c)]$, this gives \begin{align*} \bigl(\partial'_n \circ H_n(f_C)\bigr)([c]) = [f_{A,n-1}(a)]. \end{align*}[/step]

[guided]The primed connecting homomorphism is computed by differentiating the chosen lift $f_{B,n}(b) \in B'_n$. Since $f_B$ is a chain map, it commutes with differentials: \begin{align*} d_{B'}(f_{B,n}(b)) = f_{B,n-1}(d_B b). \end{align*} From the construction of $a \in A_{n-1}$, we have \begin{align*} d_B b = i_{n-1}(a). \end{align*} Substituting this into the previous identity gives \begin{align*} d_{B'}(f_{B,n}(b)) = f_{B,n-1}(i_{n-1}(a)). \end{align*} Now we use the other commutative square in the morphism of short exact sequences, namely \begin{align*} f_{B,n-1} \circ i_{n-1} = i'_{n-1} \circ f_{A,n-1}. \end{align*} Applying this identity to $a$ gives \begin{align*} f_{B,n-1}(i_{n-1}(a)) = i'_{n-1}(f_{A,n-1}(a)). \end{align*} Therefore \begin{align*} d_{B'}(f_{B,n}(b)) = i'_{n-1}(f_{A,n-1}(a)). \end{align*} This is precisely the defining situation for the primed connecting homomorphism: the cycle $f_{C,n}(c)$ is lifted by $f_{B,n}(b)$, and the differential of this lift is the image under $i'_{n-1}$ of $f_{A,n-1}(a)$. Hence \begin{align*} \partial'_n([f_{C,n}(c)]) = [f_{A,n-1}(a)]. \end{align*} Since $H_n(f_C)([c]) = [f_{C,n}(c)]$, we conclude that \begin{align*} \bigl(\partial'_n \circ H_n(f_C)\bigr)([c]) = [f_{A,n-1}(a)]. \end{align*}[/guided]

[step:Compare the two composites on every homology class] The computation of the first composite gave \begin{align*} \bigl(H_{n-1}(f_A) \circ \partial_n\bigr)([c]) = [f_{A,n-1}(a)]. \end{align*} The computation of the second composite gave \begin{align*} \bigl(\partial'_n \circ H_n(f_C)\bigr)([c]) = [f_{A,n-1}(a)]. \end{align*} Therefore \begin{align*} \bigl(H_{n-1}(f_A) \circ \partial_n\bigr)([c]) = \bigl(\partial'_n \circ H_n(f_C)\bigr)([c]). \end{align*} Since $[c] \in H_n(C_*)$ was arbitrary, the homomorphisms are equal: \begin{align*} H_{n-1}(f_A) \circ \partial_n = \partial'_n \circ H_n(f_C). \end{align*} This proves the naturality of the connecting homomorphism for every $n \in \mathbb Z$. [/step]