[proofplan] We prove commutativity square by square in the long exact sequence. The squares involving $i$ and $p$ follow directly from the commutative chain-level diagrams and the definition of induced maps on homology. The only non-formal point is the square involving the connecting homomorphism, which we verify by choosing a lift in $B_n$, applying the chain map $f_B$, and using exactness to identify the resulting boundary in $A'_{n-1}$. [/proofplan] [step:Fix the notation for cycles, boundaries, and connecting homomorphisms] For a chain complex $X_*$ with differential $d_{X,n}: X_n \to X_{n-1}$, define \begin{align*} Z_n(X_*) &:= \ker(d_{X,n}), & B_n(X_*) &:= \operatorname{im}(d_{X,n+1}), & H_n(X_*) &:= Z_n(X_*) / B_n(X_*). \end{align*} For a cycle $x \in Z_n(X_*)$, write $[x] \in H_n(X_*)$ for its homology class. The connecting homomorphism \begin{align*} \partial_n: H_n(C_*) \to H_{n-1}(A_*) \end{align*} is defined as follows. Given $[c] \in H_n(C_*)$ with $c \in Z_n(C_*)$, choose $b \in B_n$ such that $p_n(b) = c$. Since $c$ is a cycle and $p$ 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*} Exactness of \begin{align*} 0 \to A_{n-1} \xrightarrow{i_{n-1}} B_{n-1} \xrightarrow{p_{n-1}} C_{n-1} \to 0 \end{align*} gives a unique element $a \in A_{n-1}$ such that \begin{align*} i_{n-1}(a) = d_{B,n}(b). \end{align*} We first verify that $a$ is a cycle. Since $i: A_* \to B_*$ 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*} Exactness at $A_{n-2}$ gives injectivity of $i_{n-2}$, hence $d_{A,n-1}(a)=0$ and $a \in Z_{n-1}(A_*)$. Therefore $[a] \in H_{n-1}(A_*)$ is meaningful, and we set \begin{align*} \partial_n([c]) := [a]. \end{align*} This assignment is independent of the lift and of the representative. If $b_1,b_2 \in B_n$ are two lifts of the same cycle $c$, then $p_n(b_1-b_2)=0$, so exactness at $B_n$ gives $u \in A_n$ with $i_n(u)=b_1-b_2$. If $a_1,a_2 \in A_{n-1}$ are the corresponding elements, then \begin{align*} i_{n-1}(a_1-a_2) &= d_{B,n}(b_1-b_2) \\ &= d_{B,n}(i_n(u)) \\ &= i_{n-1}(d_{A,n}(u)). \end{align*} Injectivity of $i_{n-1}$ gives $a_1-a_2=d_{A,n}(u)$, so $[a_1]=[a_2]$ in $H_{n-1}(A_*)$. If $c$ is replaced by the homologous cycle $c+d_{C,n+1}(v)$ with $v \in C_{n+1}$, choose $w \in B_{n+1}$ with $p_{n+1}(w)=v$ by surjectivity of $p_{n+1}$. Then $b+d_{B,n+1}(w)$ is a lift of $c+d_{C,n+1}(v)$, and its boundary is still $d_{B,n}(b)$ because $d_{B,n}d_{B,n+1}=0$. Thus the same class $[a]$ is obtained. Hence $\partial_n: H_n(C_*) \to H_{n-1}(A_*)$ is a well-defined homomorphism. The primed connecting homomorphism $\partial'_n: H_n(C'_*) \to H_{n-1}(A'_*)$ is defined in the same way using $i'$, $p'$, and the differential of $B'_*$; the same argument proves that it is well-defined. [/step] [step:Show that the squares involving the inclusion maps commute] Let $n \in \mathbb{Z}$ and let $[a] \in H_n(A_*)$ be represented by a cycle $a \in Z_n(A_*)$. The chain-level commutativity condition is \begin{align*} f_{B,n} \circ i_n = i'_n \circ f_{A,n}. \end{align*} Therefore \begin{align*} \bigl(H_n(f_B) \circ H_n(i)\bigr)([a]) &= H_n(f_B)([i_n(a)]) \\ &= [f_{B,n}(i_n(a))] \\ &= [i'_n(f_{A,n}(a))] \\ &= H_n(i')([f_{A,n}(a)]) \\ &= \bigl(H_n(i') \circ H_n(f_A)\bigr)([a]). \end{align*} Since every element of $H_n(A_*)$ is represented by such a cycle $a$, the square involving $H_n(i)$ and $H_n(i')$ commutes. [/step] [step:Show that the squares involving the projection maps commute] Let $n \in \mathbb{Z}$ and let $[b] \in H_n(B_*)$ be represented by a cycle $b \in Z_n(B_*)$. The chain-level commutativity condition is \begin{align*} f_{C,n} \circ p_n = p'_n \circ f_{B,n}. \end{align*} Hence \begin{align*} \bigl(H_n(f_C) \circ H_n(p)\bigr)([b]) &= H_n(f_C)([p_n(b)]) \\ &= [f_{C,n}(p_n(b))] \\ &= [p'_n(f_{B,n}(b))] \\ &= H_n(p')([f_{B,n}(b)]) \\ &= \bigl(H_n(p') \circ H_n(f_B)\bigr)([b]). \end{align*} Thus the square involving $H_n(p)$ and $H_n(p')$ commutes. [/step] [step:Compare the two ways around the connecting square] Let $n \in \mathbb{Z}$ and let $[c] \in H_n(C_*)$ be represented by a cycle $c \in Z_n(C_*)$. Choose $b \in B_n$ such that \begin{align*} p_n(b) = c. \end{align*} By the definition of $\partial_n$, there is a unique element $a \in A_{n-1}$ satisfying \begin{align*} i_{n-1}(a) = d_{B,n}(b), \end{align*} and \begin{align*} \partial_n([c]) = [a]. \end{align*} Now consider the image cycle $f_{C,n}(c) \in Z_n(C'_*)$. Since the square with $p_n$ commutes, \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 valid lift of $f_{C,n}(c)$ for the definition of $\partial'_n$. Because $f_B$ is a chain map, \begin{align*} d_{B',n}(f_{B,n}(b)) = f_{B,n-1}(d_{B,n}(b)). \end{align*} Using $d_{B,n}(b) = i_{n-1}(a)$ and the commutative square with $i_{n-1}$, we get \begin{align*} d_{B',n}(f_{B,n}(b)) &= f_{B,n-1}(i_{n-1}(a)) \\ &= i'_{n-1}(f_{A,n-1}(a)). \end{align*} Therefore the definition of the primed connecting homomorphism gives \begin{align*} \partial'_n([f_{C,n}(c)]) = [f_{A,n-1}(a)]. \end{align*} Equivalently, \begin{align*} (\partial'_n \circ H_n(f_C))([c]) &= [f_{A,n-1}(a)] \\ &= H_{n-1}(f_A)([a]) \\ &= (H_{n-1}(f_A) \circ \partial_n)([c]). \end{align*} Since every element of $H_n(C_*)$ is represented by a cycle $c \in Z_n(C_*)$, the connecting square \begin{align*} H_{n-1}(f_A) \circ \partial_n = \partial'_n \circ H_n(f_C) \end{align*} commutes. [guided] We must prove that applying the connecting homomorphism first and then $H_{n-1}(f_A)$ gives the same result as applying $H_n(f_C)$ first and then the primed connecting homomorphism. Start with a class $[c] \in H_n(C_*)$, where $c \in Z_n(C_*)$. The definition of $\partial_n$ begins by lifting $c$ through $p_n$. Since the sequence \begin{align*} 0 \to A_n \xrightarrow{i_n} B_n \xrightarrow{p_n} C_n \to 0 \end{align*} is exact, $p_n$ is surjective, so there exists $b \in B_n$ such that \begin{align*} p_n(b) = c. \end{align*} Because $c$ is a cycle and $p$ 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 $d_{B,n}(b)$ lies in $\ker(p_{n-1})$. Exactness at $B_{n-1}$ says that \begin{align*} \ker(p_{n-1}) = \operatorname{im}(i_{n-1}), \end{align*} so there is a unique $a \in A_{n-1}$ with \begin{align*} i_{n-1}(a) = d_{B,n}(b). \end{align*} Before writing $[a]$, we must check 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*} Exactness at $A_{n-2}$ makes $i_{n-2}$ injective, so $d_{A,n-1}(a)=0$ and $a \in Z_{n-1}(A_*)$. The connecting homomorphism is well-defined by the lift and representative independence verified in the first step: changing the lift alters $a$ by a boundary in $A_{n-1}$, and changing $c$ by a boundary leaves the resulting homology class unchanged. Therefore \begin{align*} \partial_n([c]) = [a]. \end{align*} Now follow the other route around the square. The class $H_n(f_C)([c])$ is represented by $f_{C,n}(c) \in C'_n$. To compute $\partial'_n([f_{C,n}(c)])$, we need a lift of $f_{C,n}(c)$ through $p'_n$. The natural candidate is $f_{B,n}(b)$, and it works because the chain-level square with $p_n$ commutes: \begin{align*} p'_n(f_{B,n}(b)) = f_{C,n}(p_n(b)) = f_{C,n}(c). \end{align*} Next compute the boundary of this lift in $B'_*$: \begin{align*} d_{B',n}(f_{B,n}(b)) = f_{B,n-1}(d_{B,n}(b)), \end{align*} where we used that $f_B$ is a chain map. Since $d_{B,n}(b) = i_{n-1}(a)$, and since the square with $i_{n-1}$ commutes, this becomes \begin{align*} d_{B',n}(f_{B,n}(b)) &= f_{B,n-1}(i_{n-1}(a)) \\ &= i'_{n-1}(f_{A,n-1}(a)). \end{align*} Therefore the element of $A'_{n-1}$ selected by the definition of $\partial'_n$ is precisely $f_{A,n-1}(a)$. Hence \begin{align*} \partial'_n(H_n(f_C)([c])) &= \partial'_n([f_{C,n}(c)]) \\ &= [f_{A,n-1}(a)] \\ &= H_{n-1}(f_A)([a]) \\ &= H_{n-1}(f_A)(\partial_n([c])). \end{align*} This proves \begin{align*} \partial'_n \circ H_n(f_C) = H_{n-1}(f_A) \circ \partial_n. \end{align*} [/guided] [/step] [step:Conclude that the entire ladder of long exact sequences commutes] The [long exact homology sequence](/theorems/4210) associated to a short exact sequence of chain complexes consists, in each degree, only of maps of the forms \begin{align*} H_n(i), \qquad H_n(p), \qquad \partial_n. \end{align*} The preceding steps prove commutativity of the corresponding squares for $H_n(i)$, for $H_n(p)$, and for $\partial_n$, for every $n \in \mathbb{Z}$. Therefore every square in the ladder between the two long exact homology sequences commutes, with vertical maps $H_n(f_A)$, $H_n(f_B)$, and $H_n(f_C)$ in the appropriate degrees. This proves the asserted naturality of the long exact homology sequence. [/step]