[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 $b$.
[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]
[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 $H_n(f_C)([c]) \in H_n(C'_*)$.
[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]
[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*}
[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]
[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]
