[proofplan]
The induced maps on kernels are obtained by restriction, and the induced maps on cokernels are obtained by quotienting; the commutativity assumptions with the vertical maps are exactly what makes these operations well-defined. The only non-formal point is the square involving the connecting homomorphism. For that square, choose an element of $\ker c^D$, lift it to $B_1^D$, apply the diagram morphism, and compare the defining representative in $\operatorname{coker} a^E$ with the image of the defining representative from $\operatorname{coker} a^D$.
[/proofplan]
[step:Restrict the diagram morphism to the kernel terms]
For $X \in \{D,E\}$, write
\begin{align*}
K_A^X &:= \ker a^X \subset A_1^X, &
K_B^X &:= \ker b^X \subset B_1^X, &
K_C^X &:= \ker c^X \subset C_1^X.
\end{align*}
The commutation relations with the vertical maps imply
\begin{align*}
a^E(F_{A,1}(u)) &= F_{A,0}(a^D(u)), \\
b^E(F_{B,1}(v)) &= F_{B,0}(b^D(v)), \\
c^E(F_{C,1}(w)) &= F_{C,0}(c^D(w)).
\end{align*}
Thus, if $u \in K_A^D$, $v \in K_B^D$, and $w \in K_C^D$, then $F_{A,1}(u) \in K_A^E$, $F_{B,1}(v) \in K_B^E$, and $F_{C,1}(w) \in K_C^E$. Hence $F$ induces the restricted $R$-linear maps
\begin{align*}
\widehat{F}_{A,1} &: K_A^D \to K_A^E, &
\widehat{F}_{B,1} &: K_B^D \to K_B^E, &
\widehat{F}_{C,1} &: K_C^D \to K_C^E.
\end{align*}
[/step]
[step:Pass the diagram morphism to the cokernel terms]
For $X \in \{D,E\}$, write
\begin{align*}
Q_A^X &:= \operatorname{coker} a^X = A_0^X / a^X(A_1^X), \\
Q_B^X &:= \operatorname{coker} b^X = B_0^X / b^X(B_1^X), \\
Q_C^X &:= \operatorname{coker} c^X = C_0^X / c^X(C_1^X).
\end{align*}
Because
\begin{align*}
F_{A,0}(a^D(A_1^D)) &\subset a^E(A_1^E), \\
F_{B,0}(b^D(B_1^D)) &\subset b^E(B_1^E), \\
F_{C,0}(c^D(C_1^D)) &\subset c^E(C_1^E),
\end{align*}
the maps $F_{A,0}$, $F_{B,0}$, and $F_{C,0}$ descend to quotient maps
\begin{align*}
\overline{F}_{A,0} &: Q_A^D \to Q_A^E, &
\overline{F}_{B,0} &: Q_B^D \to Q_B^E, &
\overline{F}_{C,0} &: Q_C^D \to Q_C^E,
\end{align*}
given by
\begin{align*}
\overline{F}_{A,0}([u]) &= [F_{A,0}(u)], \\
\overline{F}_{B,0}([v]) &= [F_{B,0}(v)], \\
\overline{F}_{C,0}([w]) &= [F_{C,0}(w)].
\end{align*}
Here $[u]$, $[v]$, and $[w]$ denote residue classes in the appropriate cokernel modules.
[/step]
[step:Verify commutativity away from the connecting homomorphism]
The maps in the [snake lemma](/theorems/1930) exact sequence before $\delta_X$ are the restrictions of $i_1^X$ and $p_1^X$ to the corresponding kernels. The horizontal commutation relations
\begin{align*}
F_{B,1} \circ i_1^D &= i_1^E \circ F_{A,1}, &
F_{C,1} \circ p_1^D &= p_1^E \circ F_{B,1}
\end{align*}
therefore restrict to
\begin{align*}
\widehat{F}_{B,1} \circ i_1^D\big|_{K_A^D}
&=
i_1^E\big|_{K_A^E} \circ \widehat{F}_{A,1}, \\
\widehat{F}_{C,1} \circ p_1^D\big|_{K_B^D}
&=
p_1^E\big|_{K_B^E} \circ \widehat{F}_{B,1}.
\end{align*}
The maps in the [snake lemma](/theorems/4533) exact sequence after $\delta_X$ are induced by $i_0^X$ and $p_0^X$ on cokernels. The horizontal commutation relations
\begin{align*}
F_{B,0} \circ i_0^D &= i_0^E \circ F_{A,0}, &
F_{C,0} \circ p_0^D &= p_0^E \circ F_{B,0}
\end{align*}
give, for every $[u] \in Q_A^D$ and $[v] \in Q_B^D$,
\begin{align*}
\overline{F}_{B,0}([i_0^D(u)])
&=
[F_{B,0}(i_0^D(u))]
=
[i_0^E(F_{A,0}(u))]
=
i_0^E{}_{\operatorname{coker}}(\overline{F}_{A,0}([u])), \\
\overline{F}_{C,0}([p_0^D(v)])
&=
[F_{C,0}(p_0^D(v))]
=
[p_0^E(F_{B,0}(v))]
=
p_0^E{}_{\operatorname{coker}}(\overline{F}_{B,0}([v])).
\end{align*}
Thus all squares not involving $\delta_D$ and $\delta_E$ commute.
[/step]
[step:Compare the two constructions of the connecting class]
Let $x \in K_C^D = \ker c^D$. Since $p_1^D: B_1^D \to C_1^D$ is surjective, choose $y \in B_1^D$ such that $p_1^D(y)=x$. The defining construction of the connecting homomorphism uses the element $z \in A_0^D$ uniquely determined by
\begin{align*}
i_0^D(z)=b^D(y).
\end{align*}
Existence follows because
\begin{align*}
p_0^D(b^D(y))=c^D(p_1^D(y))=c^D(x)=0,
\end{align*}
so $b^D(y) \in \ker p_0^D=\operatorname{im} i_0^D$; uniqueness follows because $i_0^D$ is injective. Therefore
\begin{align*}
\delta_D(x)=[z] \in Q_A^D.
\end{align*}
Now apply the morphism of diagrams. The element $F_{B,1}(y) \in B_1^E$ is a lift of $F_{C,1}(x) \in C_1^E$, since
\begin{align*}
p_1^E(F_{B,1}(y))
=
F_{C,1}(p_1^D(y))
=
F_{C,1}(x).
\end{align*}
Moreover $F_{C,1}(x) \in \ker c^E$ by the first step. The defining representative for $\delta_E(F_{C,1}(x))$ is the unique element $z_E \in A_0^E$ satisfying
\begin{align*}
i_0^E(z_E)=b^E(F_{B,1}(y)).
\end{align*}
Using commutativity with $b$ and then with $i_0$, we compute
\begin{align*}
b^E(F_{B,1}(y))
=
F_{B,0}(b^D(y))
=
F_{B,0}(i_0^D(z))
=
i_0^E(F_{A,0}(z)).
\end{align*}
Since $i_0^E$ is injective, $z_E=F_{A,0}(z)$. Hence
\begin{align*}
\delta_E(F_{C,1}(x))
=
[F_{A,0}(z)]
=
\overline{F}_{A,0}([z])
=
\overline{F}_{A,0}(\delta_D(x)).
\end{align*}
Because $x \in \ker c^D$ was arbitrary,
\begin{align*}
\delta_E \circ \widehat{F}_{C,1}
=
\overline{F}_{A,0} \circ \delta_D.
\end{align*}
[guided]
We prove the naturality square for the connecting homomorphism directly from the construction of the snake map. Take an arbitrary element $x \in K_C^D=\ker c^D$. The element $x$ lies in the right-hand kernel of the diagram $D$. Since the top row of $D$ is exact, the map $p_1^D: B_1^D \to C_1^D$ is surjective, so there is an element $y \in B_1^D$ with
\begin{align*}
p_1^D(y)=x.
\end{align*}
The snake construction now sends $y$ downward by $b^D$. This element lands in $\ker p_0^D$ because the right square of $D$ commutes:
\begin{align*}
p_0^D(b^D(y))
=
c^D(p_1^D(y))
=
c^D(x)
=
0.
\end{align*}
Exactness of the bottom row says $\ker p_0^D=\operatorname{im} i_0^D$, so there exists $z \in A_0^D$ such that
\begin{align*}
i_0^D(z)=b^D(y).
\end{align*}
The element $z$ is unique because $i_0^D$ is injective. Thus the connecting homomorphism for $D$ is
\begin{align*}
\delta_D(x)=[z] \in A_0^D/a^D(A_1^D).
\end{align*}
Now compare this with the same construction after applying the morphism $F: D \to E$. The element $F_{B,1}(y) \in B_1^E$ is a lift of $F_{C,1}(x) \in C_1^E$, since the top horizontal square commutes:
\begin{align*}
p_1^E(F_{B,1}(y))
=
F_{C,1}(p_1^D(y))
=
F_{C,1}(x).
\end{align*}
Also $F_{C,1}(x) \in \ker c^E$ because
\begin{align*}
c^E(F_{C,1}(x))
=
F_{C,0}(c^D(x))
=
F_{C,0}(0)
=
0.
\end{align*}
The connecting homomorphism $\delta_E$ therefore computes its class using the unique element $z_E \in A_0^E$ satisfying
\begin{align*}
i_0^E(z_E)=b^E(F_{B,1}(y)).
\end{align*}
We identify this $z_E$. First commute $F$ with the middle vertical map, then commute $F$ with the lower left horizontal map:
\begin{align*}
b^E(F_{B,1}(y))
=
F_{B,0}(b^D(y))
=
F_{B,0}(i_0^D(z))
=
i_0^E(F_{A,0}(z)).
\end{align*}
Both $z_E$ and $F_{A,0}(z)$ map under the injective map $i_0^E$ to the same element of $B_0^E$. Hence
\begin{align*}
z_E=F_{A,0}(z).
\end{align*}
Therefore
\begin{align*}
\delta_E(F_{C,1}(x))
=
[F_{A,0}(z)]
=
\overline{F}_{A,0}([z])
=
\overline{F}_{A,0}(\delta_D(x)).
\end{align*}
Since this equality holds for every $x \in \ker c^D$, the square involving the connecting homomorphisms commutes:
\begin{align*}
\delta_E \circ \widehat{F}_{C,1}
=
\overline{F}_{A,0} \circ \delta_D.
\end{align*}
[/guided]
[/step]
[step:Assemble the commuting morphism of exact sequences]
The preceding steps construct the induced maps on all six terms of the two snake lemma exact sequences and verify commutativity with each arrow. Hence the diagram
\begin{align*}
\ker a^D \longrightarrow \ker b^D \longrightarrow \ker c^D
\xrightarrow{\delta_D}
\operatorname{coker} a^D \longrightarrow \operatorname{coker} b^D \longrightarrow \operatorname{coker} c^D
\end{align*}
maps to
\begin{align*}
\ker a^E \longrightarrow \ker b^E \longrightarrow \ker c^E
\xrightarrow{\delta_E}
\operatorname{coker} a^E \longrightarrow \operatorname{coker} b^E \longrightarrow \operatorname{coker} c^E
\end{align*}
as a commutative diagram of $R$-modules. In particular,
\begin{align*}
\overline{F}_{A,0} \circ \delta_D
=
\delta_E \circ F_{C,1}\big|_{\ker c^D},
\end{align*}
which is the [naturality of the connecting homomorphism](/theorems/4538). This proves the theorem.
[/step]
