[proofplan] We construct the connecting homomorphism $\partial_*: H_n(C_\bullet) \to H_{n-1}(A_\bullet)$ by a diagram chase through the short exact sequence $0 \to A_n \xrightarrow{i_n} B_n \xrightarrow{q_n} C_n \to 0$. Given a cycle $x \in Z_n(C)$, we lift it to $B_n$ via surjectivity of $q_n$, apply the boundary $d^B$, and use exactness to land in $A_{n-1}$. We verify the result is a cycle in $A_{n-1}$, show independence of the lift, and prove exactness at each of the three positions in the long exact sequence. [/proofplan] [step:Construct the connecting homomorphism $\partial_*$] Fix the commutative diagram with exact rows: \begin{align*} 0 \to A_n \xrightarrow{i_n} B_n \xrightarrow{q_n} C_n \to 0 \\ \quad\;\; \downarrow d^A \qquad \downarrow d^B \qquad \downarrow d^C \\ 0 \to A_{n-1} \xrightarrow{i_{n-1}} B_{n-1} \xrightarrow{q_{n-1}} C_{n-1} \to 0 \end{align*} Let $[x] \in H_n(C_\bullet)$ with $x \in Z_n(C)$, so $d^C_n(x) = 0$. Since $q_n$ is surjective, there exists $y \in B_n$ with $q_n(y) = x$. We claim $d^B_n(y) \in \operatorname{im}(i_{n-1})$. Indeed: \begin{align*} q_{n-1}(d^B_n(y)) = d^C_n(q_n(y)) = d^C_n(x) = 0, \end{align*} where the first equality uses the commutativity $q_{n-1} \circ d^B_n = d^C_n \circ q_n$ (which holds because $q_\bullet$ is a chain map). Since $q_{n-1}(d^B_n(y)) = 0$ and the row is exact at $B_{n-1}$ (i.e., $\ker(q_{n-1}) = \operatorname{im}(i_{n-1})$), there exists a unique $a \in A_{n-1}$ with $i_{n-1}(a) = d^B_n(y)$. Uniqueness follows from the injectivity of $i_{n-1}$. Define $\partial_*([x]) = [a] \in H_{n-1}(A_\bullet)$. [guided] The construction follows the only path available through the diagram. Starting with a cycle $x \in C_n$: 1. **Lift:** Since $q_n$ is surjective, choose $y \in B_n$ mapping to $x$. 2. **Apply boundary:** Compute $d^B_n(y) \in B_{n-1}$. 3. **Land in $A$:** Show $d^B_n(y) \in \ker(q_{n-1}) = \operatorname{im}(i_{n-1})$, so $d^B_n(y) = i_{n-1}(a)$ for a unique $a$. 4. **Declare:** $\partial_*([x]) = [a]$. Why does $d^B_n(y)$ land in $\ker(q_{n-1})$? Because $q_{n-1}(d^B_n(y)) = d^C_n(q_n(y)) = d^C_n(x) = 0$, using commutativity and the fact that $x$ is a cycle. For this to define a map on homology, we must verify: (i) $a$ is a cycle in $A_{n-1}$, and (ii) the class $[a]$ does not depend on the choice of lift $y$. [/guided] [/step] [step:Verify that $a$ is a cycle in $A_{n-1}$] We must show $d^A_{n-1}(a) = 0$. Since $i_{n-2}$ is injective, it suffices to show $i_{n-2}(d^A_{n-1}(a)) = 0$. By commutativity ($i_{n-2} \circ d^A_{n-1} = d^B_{n-1} \circ i_{n-1}$, since $i_\bullet$ 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(y)) = 0, \end{align*} where the last equality uses $d^B_{n-1} \circ d^B_n = 0$ (the chain complex condition for $B_\bullet$). Since $i_{n-2}$ is injective, $d^A_{n-1}(a) = 0$, so $a \in Z_{n-1}(A)$. [/step] [step:Show $\partial_*$ is independent of the choice of lift] Suppose $y' \in B_n$ is another lift with $q_n(y') = x$. Then $q_n(y - y') = 0$, so $y - y' \in \ker(q_n) = \operatorname{im}(i_n)$. Write $y - y' = i_n(w)$ for some $w \in A_n$. The corresponding elements $a, a' \in A_{n-1}$ satisfy \begin{align*} i_{n-1}(a - a') = d^B_n(y) - d^B_n(y') = d^B_n(y - y') = d^B_n(i_n(w)) = i_{n-1}(d^A_n(w)), \end{align*} where the last equality uses commutativity $d^B_n \circ i_n = i_{n-1} \circ d^A_n$. By injectivity of $i_{n-1}$, $a - a' = d^A_n(w)$, so $[a] = [a']$ in $H_{n-1}(A_\bullet)$. [/step] [step:Show $\partial_*$ is independent of the representative of $[x]$] Suppose $[x] = [x']$, so $x - x' = d^C_{n+1}(z)$ for some $z \in C_{n+1}$. Let $y, y'$ be lifts of $x, x'$ respectively. Since $q_{n+1}$ is surjective, choose $w \in B_{n+1}$ with $q_{n+1}(w) = z$. Then \begin{align*} q_n(y - y' - d^B_{n+1}(w)) = x - x' - d^C_{n+1}(q_{n+1}(w)) = x - x' - d^C_{n+1}(z) = 0. \end{align*} So $y - y' - d^B_{n+1}(w) \in \ker(q_n) = \operatorname{im}(i_n)$. Write $y - y' - d^B_{n+1}(w) = i_n(v)$ for some $v \in A_n$. Then \begin{align*} i_{n-1}(a - a') = d^B_n(y - y') = d^B_n(i_n(v) + d^B_{n+1}(w)) = i_{n-1}(d^A_n(v)) + 0, \end{align*} using $d^B_n \circ d^B_{n+1} = 0$. By injectivity of $i_{n-1}$, $a - a' = d^A_n(v)$, so $[a] = [a']$. Therefore $\partial_*$ is well-defined on $H_n(C_\bullet)$. [guided] This step addresses the question: what if we change the cycle $x$ to a homologous cycle $x' = x - d^C(z)$? The lifts of $x$ and $x'$ may differ, and we need to show the resulting elements of $H_{n-1}(A)$ agree. The idea is to lift the boundary $d^C(z)$ as well: lift $z$ to $w$ in $B_{n+1}$, then $d^B(w)$ lifts $d^C(z)$. Adjusting the lift of $x$ by $d^B(w)$ gives a lift of $x'$, and the difference $y - y' - d^B(w)$ lies in $\ker(q_n) = \operatorname{im}(i_n)$. Applying $d^B$ and using $d^B \circ d^B = 0$ shows the two elements $a$ and $a'$ differ by a boundary in $A_\bullet$. [/guided] [/step] [step:Verify that $\partial_*$ is a homomorphism] Let $[x_1], [x_2] \in H_n(C_\bullet)$ with lifts $y_1, y_2 \in B_n$. Then $y_1 + y_2$ is a lift of $x_1 + x_2$, and $d^B_n(y_1 + y_2) = d^B_n(y_1) + d^B_n(y_2) = i_{n-1}(a_1) + i_{n-1}(a_2) = i_{n-1}(a_1 + a_2)$. Therefore $\partial_*([x_1 + x_2]) = [a_1 + a_2] = [a_1] + [a_2] = \partial_*([x_1]) + \partial_*([x_2])$. [/step] [step:Prove exactness at $H_n(B)$: $\ker(q_*) = \operatorname{im}(i_*)$] **$\operatorname{im}(i_*) \subset \ker(q_*)$:** For any $[\alpha] \in H_n(A)$, $q_*(i_*([\alpha])) = [(q_n \circ i_n)(\alpha)] = [0] = 0$, since $q_n \circ i_n = 0$ by exactness of the row at $B_n$. **$\ker(q_*) \subset \operatorname{im}(i_*)$:** Suppose $[b] \in H_n(B)$ with $q_*([b]) = 0$, i.e., $q_n(b) = d^C_{n+1}(z)$ for some $z \in C_{n+1}$. Lift $z$ to $w \in B_{n+1}$ with $q_{n+1}(w) = z$. Then $q_n(b - d^B_{n+1}(w)) = q_n(b) - d^C_{n+1}(q_{n+1}(w)) = q_n(b) - d^C_{n+1}(z) = 0$. By exactness of the row at $B_n$, $b - d^B_{n+1}(w) = i_n(\alpha)$ for some $\alpha \in A_n$. We verify $\alpha$ is a cycle: $i_{n-1}(d^A_n(\alpha)) = d^B_n(i_n(\alpha)) = d^B_n(b - d^B_{n+1}(w)) = d^B_n(b) - 0 = 0$ (since $b$ is a cycle). By injectivity of $i_{n-1}$, $d^A_n(\alpha) = 0$. Therefore $[b] = [i_n(\alpha) + d^B_{n+1}(w)] = [i_n(\alpha)] = i_*([\alpha])$. [/step] [step:Prove exactness at $H_n(C)$: $\ker(\partial_*) = \operatorname{im}(q_*)$] **$\operatorname{im}(q_*) \subset \ker(\partial_*)$:** Let $[b] \in H_n(B)$. Then $q_*([b]) = [q_n(b)]$, and $b$ itself is a lift of $q_n(b)$. Since $b$ is a cycle, $d^B_n(b) = 0 = i_{n-1}(0)$, so $\partial_*([q_n(b)]) = [0] = 0$. **$\ker(\partial_*) \subset \operatorname{im}(q_*)$:** Suppose $\partial_*([x]) = 0$, meaning $a = d^A_n(w)$ for some $w \in A_n$ (where $a$ is defined by $i_{n-1}(a) = d^B_n(y)$ with $q_n(y) = x$). Set $y' = y - i_n(w)$. Then: - $q_n(y') = q_n(y) - q_n(i_n(w)) = x - 0 = x$, so $y'$ also lifts $x$. - $d^B_n(y') = d^B_n(y) - d^B_n(i_n(w)) = i_{n-1}(a) - i_{n-1}(d^A_n(w)) = i_{n-1}(a - a) = 0$. So $y'$ is a cycle in $B_n$ with $q_n(y') = x$, giving $q_*([y']) = [x]$ and hence $[x] \in \operatorname{im}(q_*)$. [/step] [step:Prove exactness at $H_{n-1}(A)$: $\ker(i_*) = \operatorname{im}(\partial_*)$] **$\operatorname{im}(\partial_*) \subset \ker(i_*)$:** Let $[x] \in H_n(C)$ with lift $y$ and $i_{n-1}(a) = d^B_n(y)$. Then $i_*([ a]) = [i_{n-1}(a)] = [d^B_n(y)] = 0$ in $H_{n-1}(B)$ (since $d^B_n(y)$ is a boundary). **$\ker(i_*) \subset \operatorname{im}(\partial_*)$:** Suppose $[\alpha] \in H_{n-1}(A)$ with $i_*([\alpha]) = 0$, meaning $i_{n-1}(\alpha) = d^B_n(y)$ for some $y \in B_n$. Set $x = q_n(y)$. Then $x$ is a cycle: \begin{align*} d^C_n(x) = d^C_n(q_n(y)) = q_{n-1}(d^B_n(y)) = q_{n-1}(i_{n-1}(\alpha)) = 0, \end{align*} since $q_{n-1} \circ i_{n-1} = 0$. The element $y$ lifts $x$, and $d^B_n(y) = i_{n-1}(\alpha)$, so $\partial_*([x]) = [\alpha]$ by construction. [/step]