[proofplan] We construct the connecting homomorphism $\delta$ by lifting an element of $\ker\gamma$ to $B_1$, applying $\beta$, and then using exactness of the bottom row to descend uniquely to $A_0$ modulo $\operatorname{im}\alpha$. After proving that this construction is independent of choices and is $R$-linear, we verify exactness at each term. The first and last exactness statements come directly from injectivity of $i_1$ and surjectivity of $p_0$; the middle exactness statements are precisely the compatibility conditions in the commutative diagram. [/proofplan] [step:Define the induced maps on kernels and cokernels] The restrictions \begin{align*} i_1|_{\ker\alpha}: \ker\alpha &\longrightarrow \ker\beta, \\ a_1 &\longmapsto i_1(a_1), \end{align*} and \begin{align*} p_1|_{\ker\beta}: \ker\beta &\longrightarrow \ker\gamma, \\ b_1 &\longmapsto p_1(b_1), \end{align*} are well-defined. Indeed, if $a_1 \in \ker\alpha$, then \begin{align*} \beta(i_1(a_1)) = i_0(\alpha(a_1)) = i_0(0) = 0, \end{align*} so $i_1(a_1) \in \ker\beta$. If $b_1 \in \ker\beta$, then \begin{align*} \gamma(p_1(b_1)) = p_0(\beta(b_1)) = p_0(0) = 0, \end{align*} so $p_1(b_1) \in \ker\gamma$. Define \begin{align*} \overline{i}_0: \operatorname{coker}\alpha &\longrightarrow \operatorname{coker}\beta, \\ a_0+\operatorname{im}\alpha &\longmapsto i_0(a_0)+\operatorname{im}\beta, \end{align*} and \begin{align*} \overline{p}_0: \operatorname{coker}\beta &\longrightarrow \operatorname{coker}\gamma, \\ b_0+\operatorname{im}\beta &\longmapsto p_0(b_0)+\operatorname{im}\gamma. \end{align*} These maps are well-defined. If $a_0-a_0'=\alpha(a_1)$ for some $a_1\in A_1$, then \begin{align*} i_0(a_0)-i_0(a_0') = i_0(\alpha(a_1)) = \beta(i_1(a_1)) \in \operatorname{im}\beta. \end{align*} If $b_0-b_0'=\beta(b_1)$ for some $b_1\in B_1$, then \begin{align*} p_0(b_0)-p_0(b_0') = p_0(\beta(b_1)) = \gamma(p_1(b_1)) \in \operatorname{im}\gamma. \end{align*} Thus both maps descend to cokernels. They are $R$-linear because $i_0$ and $p_0$ are $R$-linear. [/step] [step:Construct the connecting homomorphism by lifting and descending] Let $c_1 \in \ker\gamma$. Since $p_1: B_1 \to C_1$ is surjective, choose $b_1 \in B_1$ such that $p_1(b_1)=c_1$. Then \begin{align*} p_0(\beta(b_1)) = \gamma(p_1(b_1)) = \gamma(c_1)=0. \end{align*} Hence $\beta(b_1)\in\ker p_0$. Exactness of the bottom row gives $\ker p_0=\operatorname{im}i_0$, so there exists $a_0\in A_0$ such that \begin{align*} i_0(a_0)=\beta(b_1). \end{align*} Define \begin{align*} \delta: \ker\gamma &\longrightarrow \operatorname{coker}\alpha, \\ c_1 &\longmapsto a_0+\operatorname{im}\alpha. \end{align*} We prove that this definition is independent of the chosen lift $b_1$ and the chosen element $a_0$. The element $a_0$ is unique for a fixed $b_1$ because $i_0$ is injective. If $b_1'\in B_1$ is another lift of $c_1$, then $p_1(b_1-b_1')=0$, so exactness of the top row gives $b_1-b_1'\in\operatorname{im}i_1$. Choose $a_1\in A_1$ such that \begin{align*} i_1(a_1)=b_1-b_1'. \end{align*} If $a_0,a_0'\in A_0$ satisfy \begin{align*} i_0(a_0)=\beta(b_1), \qquad i_0(a_0')=\beta(b_1'), \end{align*} then \begin{align*} i_0(a_0-a_0') &= \beta(b_1)-\beta(b_1') \\ &= \beta(b_1-b_1') \\ &= \beta(i_1(a_1)) \\ &= i_0(\alpha(a_1)). \end{align*} Since $i_0$ is injective, \begin{align*} a_0-a_0'=\alpha(a_1)\in\operatorname{im}\alpha. \end{align*} Thus $a_0+\operatorname{im}\alpha=a_0'+\operatorname{im}\alpha$, so $\delta$ is well-defined. Finally, $\delta$ is $R$-linear. For $c_1,c_1'\in\ker\gamma$ and $r\in R$, choose lifts $b_1,b_1'\in B_1$ with $p_1(b_1)=c_1$ and $p_1(b_1')=c_1'$. Choose $a_0,a_0'\in A_0$ with $i_0(a_0)=\beta(b_1)$ and $i_0(a_0')=\beta(b_1')$. Then $b_1+b_1'$ lifts $c_1+c_1'$ and $r b_1$ lifts $r c_1$, while \begin{align*} i_0(a_0+a_0') &= \beta(b_1+b_1'), \\ i_0(r a_0) &= \beta(r b_1). \end{align*} Therefore \begin{align*} \delta(c_1+c_1') &= \delta(c_1)+\delta(c_1'), \\ \delta(r c_1) &= r\delta(c_1). \end{align*} [/step] [step:Prove exactness at $\ker\alpha$ and $\ker\beta$] The map $0\to\ker\alpha$ is injective by definition. Since $i_1$ is injective, its restriction $i_1|_{\ker\alpha}: \ker\alpha\to\ker\beta$ is injective, so the sequence is exact at $\ker\alpha$. Next we prove exactness at $\ker\beta$. Since $p_1\circ i_1=0$ by exactness of the top row, we have \begin{align*} \operatorname{im}(i_1|_{\ker\alpha})\subseteq \ker(p_1|_{\ker\beta}). \end{align*} Conversely, let $b_1\in\ker\beta$ satisfy $p_1(b_1)=0$. Exactness of the top row gives $b_1\in\operatorname{im}i_1$, so choose $a_1\in A_1$ with $i_1(a_1)=b_1$. Then \begin{align*} i_0(\alpha(a_1))=\beta(i_1(a_1))=\beta(b_1)=0. \end{align*} Since $i_0$ is injective, $\alpha(a_1)=0$, so $a_1\in\ker\alpha$. Hence $b_1=i_1(a_1)$ lies in $\operatorname{im}(i_1|_{\ker\alpha})$. Therefore \begin{align*} \operatorname{im}(i_1|_{\ker\alpha})=\ker(p_1|_{\ker\beta}). \end{align*} [/step] [step:Prove exactness at $\ker\gamma$] First let $b_1\in\ker\beta$. Then $p_1(b_1)\in\ker\gamma$, and in the construction of $\delta(p_1(b_1))$ we may choose the lift $b_1$ itself. Since $\beta(b_1)=0=i_0(0)$, the descended element is $0\in A_0$. Hence \begin{align*} \delta(p_1(b_1))=0+\operatorname{im}\alpha, \end{align*} so \begin{align*} \operatorname{im}(p_1|_{\ker\beta})\subseteq \ker\delta. \end{align*} Conversely, let $c_1\in\ker\gamma$ satisfy $\delta(c_1)=0$. Choose $b_1\in B_1$ with $p_1(b_1)=c_1$, and choose $a_0\in A_0$ with $i_0(a_0)=\beta(b_1)$. The equality $\delta(c_1)=0$ in $\operatorname{coker}\alpha$ means $a_0\in\operatorname{im}\alpha$, so choose $a_1\in A_1$ with $\alpha(a_1)=a_0$. Define \begin{align*} b_1' := b_1-i_1(a_1)\in B_1. \end{align*} Then \begin{align*} \beta(b_1') &= \beta(b_1)-\beta(i_1(a_1)) \\ &= i_0(a_0)-i_0(\alpha(a_1)) \\ &= 0, \end{align*} so $b_1'\in\ker\beta$. Also \begin{align*} p_1(b_1')=p_1(b_1)-p_1(i_1(a_1))=c_1-0=c_1. \end{align*} Thus $c_1\in\operatorname{im}(p_1|_{\ker\beta})$. Therefore \begin{align*} \operatorname{im}(p_1|_{\ker\beta})=\ker\delta. \end{align*} [/step] [step:Prove exactness at $\operatorname{coker}\alpha$] Let $c_1\in\ker\gamma$, and write $\delta(c_1)=a_0+\operatorname{im}\alpha$. By construction, there exists $b_1\in B_1$ such that \begin{align*} p_1(b_1)=c_1, \qquad i_0(a_0)=\beta(b_1). \end{align*} Therefore \begin{align*} \overline{i}_0(\delta(c_1)) = i_0(a_0)+\operatorname{im}\beta = \beta(b_1)+\operatorname{im}\beta = 0 \end{align*} in $\operatorname{coker}\beta$. Hence \begin{align*} \operatorname{im}\delta\subseteq\ker\overline{i}_0. \end{align*} Conversely, let $a_0+\operatorname{im}\alpha\in\ker\overline{i}_0$. Then \begin{align*} i_0(a_0)+\operatorname{im}\beta=0 \end{align*} in $\operatorname{coker}\beta$, so choose $b_1\in B_1$ such that \begin{align*} \beta(b_1)=i_0(a_0). \end{align*} Define $c_1:=p_1(b_1)\in C_1$. Then \begin{align*} \gamma(c_1)=\gamma(p_1(b_1))=p_0(\beta(b_1))=p_0(i_0(a_0))=0, \end{align*} because $p_0\circ i_0=0$ by exactness of the bottom row. Thus $c_1\in\ker\gamma$. In the construction of $\delta(c_1)$, using the lift $b_1$ gives the descended element $a_0$, so \begin{align*} \delta(c_1)=a_0+\operatorname{im}\alpha. \end{align*} Therefore \begin{align*} \ker\overline{i}_0\subseteq\operatorname{im}\delta. \end{align*} Thus \begin{align*} \operatorname{im}\delta=\ker\overline{i}_0. \end{align*} [/step] [step:Prove exactness at $\operatorname{coker}\beta$ and $\operatorname{coker}\gamma$] First, \begin{align*} \overline{p}_0(\overline{i}_0(a_0+\operatorname{im}\alpha)) = \overline{p}_0(i_0(a_0)+\operatorname{im}\beta) = p_0(i_0(a_0))+\operatorname{im}\gamma = 0 \end{align*} for every $a_0\in A_0$, so \begin{align*} \operatorname{im}\overline{i}_0\subseteq\ker\overline{p}_0. \end{align*} Conversely, let $b_0+\operatorname{im}\beta\in\ker\overline{p}_0$. Then \begin{align*} p_0(b_0)+\operatorname{im}\gamma=0 \end{align*} in $\operatorname{coker}\gamma$, so choose $c_1\in C_1$ such that \begin{align*} \gamma(c_1)=p_0(b_0). \end{align*} Since $p_1$ is surjective, choose $b_1\in B_1$ with $p_1(b_1)=c_1$. Then \begin{align*} p_0(b_0-\beta(b_1)) = p_0(b_0)-p_0(\beta(b_1)) = \gamma(c_1)-\gamma(p_1(b_1)) = 0. \end{align*} Hence $b_0-\beta(b_1)\in\ker p_0=\operatorname{im}i_0$. Choose $a_0\in A_0$ such that \begin{align*} i_0(a_0)=b_0-\beta(b_1). \end{align*} Then \begin{align*} b_0+\operatorname{im}\beta = i_0(a_0)+\operatorname{im}\beta = \overline{i}_0(a_0+\operatorname{im}\alpha). \end{align*} Thus \begin{align*} \ker\overline{p}_0\subseteq\operatorname{im}\overline{i}_0, \end{align*} and therefore \begin{align*} \operatorname{im}\overline{i}_0=\ker\overline{p}_0. \end{align*} Finally, $\overline{p}_0$ is surjective. Let $c_0+\operatorname{im}\gamma\in\operatorname{coker}\gamma$. Since $p_0$ is surjective, choose $b_0\in B_0$ such that $p_0(b_0)=c_0$. Then \begin{align*} \overline{p}_0(b_0+\operatorname{im}\beta)=c_0+\operatorname{im}\gamma. \end{align*} So the map $\operatorname{coker}\beta\to\operatorname{coker}\gamma$ is surjective, which is exactness at $\operatorname{coker}\gamma$ and at the terminal $0$. Combining the preceding steps proves exactness of the entire sequence. [/step]