[proofplan] We first verify that the construction of $\delta(c_1)$ is possible: a lift exists by surjectivity of $p_1$, and the corresponding element of $A_0$ exists and is unique by exactness of the lower short exact sequence. We then show that replacing the chosen lift changes the resulting element of $A_0$ by an element of $\operatorname{im}\alpha$, so the class in $\operatorname{coker}\alpha$ is independent of choices. Finally, we choose compatible lifts for sums and scalar multiples to prove additivity and $R$-linearity. [/proofplan] [step:Construct the class associated to an element of $\ker\gamma$] Fix $c_1 \in \ker\gamma$. Since $p_1: B_1 \to C_1$ is surjective, choose $b_1 \in B_1$ such that $p_1(b_1)=c_1$. The commutativity relation $\gamma \circ p_1=p_0\circ\beta$ gives \begin{align*} p_0(\beta(b_1))=(p_0\circ\beta)(b_1)=(\gamma\circ p_1)(b_1)=\gamma(c_1)=0. \end{align*} Thus $\beta(b_1)\in\ker p_0$. Exactness of the lower sequence gives $\ker p_0=\operatorname{im} i_0$, so there exists $a_0\in A_0$ such that $i_0(a_0)=\beta(b_1)$. Since $i_0$ is injective, this $a_0$ is unique. Hence the formula \begin{align*} \delta(c_1):=a_0+\operatorname{im}\alpha \end{align*} is meaningful once the lift $b_1$ has been chosen. [/step] [step:Show that changing the lift changes $a_0$ by an element of $\operatorname{im}\alpha$] Let $b_{1,0},b_{1,1}\in B_1$ be two lifts of the same element $c_1\in\ker\gamma$, so \begin{align*} p_1(b_{1,0})=c_1=p_1(b_{1,1}). \end{align*} Let $a_{0,0},a_{0,1}\in A_0$ be the unique elements satisfying \begin{align*} i_0(a_{0,0})=\beta(b_{1,0}), \qquad i_0(a_{0,1})=\beta(b_{1,1}). \end{align*} Then \begin{align*} p_1(b_{1,0}-b_{1,1})=p_1(b_{1,0})-p_1(b_{1,1})=0, \end{align*} so $b_{1,0}-b_{1,1}\in\ker p_1$. Exactness of the upper sequence gives $\ker p_1=\operatorname{im}i_1$, hence there exists $a_1\in A_1$ such that \begin{align*} b_{1,0}-b_{1,1}=i_1(a_1). \end{align*} Applying $\beta$ and using $\beta\circ i_1=i_0\circ\alpha$, we obtain \begin{align*} i_0(a_{0,0}-a_{0,1}) &=\beta(b_{1,0})-\beta(b_{1,1}) \\ &=\beta(b_{1,0}-b_{1,1}) \\ &=\beta(i_1(a_1)) \\ &=i_0(\alpha(a_1)). \end{align*} Since $i_0$ is injective, \begin{align*} a_{0,0}-a_{0,1}=\alpha(a_1)\in\operatorname{im}\alpha. \end{align*} Therefore \begin{align*} a_{0,0}+\operatorname{im}\alpha=a_{0,1}+\operatorname{im}\alpha. \end{align*} Thus $\delta(c_1)$ is independent of the chosen lift. [guided] The only possible ambiguity in the definition of $\delta(c_1)$ is the choice of the lift $b_1\in B_1$. We must prove that two different choices produce the same coset in $A_0/\operatorname{im}\alpha$. Let $b_{1,0},b_{1,1}\in B_1$ be two lifts of $c_1$, meaning \begin{align*} p_1(b_{1,0})=c_1=p_1(b_{1,1}). \end{align*} Let $a_{0,0},a_{0,1}\in A_0$ be the uniquely determined elements such that \begin{align*} i_0(a_{0,0})=\beta(b_{1,0}), \qquad i_0(a_{0,1})=\beta(b_{1,1}). \end{align*} The difference of the two lifts lies in the kernel of $p_1$ because \begin{align*} p_1(b_{1,0}-b_{1,1})=p_1(b_{1,0})-p_1(b_{1,1})=c_1-c_1=0. \end{align*} By exactness of the upper short exact sequence, $\ker p_1=\operatorname{im}i_1$. Hence there is an element $a_1\in A_1$ satisfying \begin{align*} b_{1,0}-b_{1,1}=i_1(a_1). \end{align*} Now apply $\beta$ to this equality. Since the left square commutes, $\beta\circ i_1=i_0\circ\alpha$, so \begin{align*} i_0(a_{0,0}-a_{0,1}) &=\beta(b_{1,0})-\beta(b_{1,1}) \\ &=\beta(b_{1,0}-b_{1,1}) \\ &=\beta(i_1(a_1)) \\ &=i_0(\alpha(a_1)). \end{align*} The map $i_0$ is injective because the lower sequence is exact at $A_0$. Therefore \begin{align*} a_{0,0}-a_{0,1}=\alpha(a_1). \end{align*} This says exactly that the two possible representatives $a_{0,0}$ and $a_{0,1}$ differ by an element of $\operatorname{im}\alpha$. Hence they define the same coset: \begin{align*} a_{0,0}+\operatorname{im}\alpha=a_{0,1}+\operatorname{im}\alpha. \end{align*} Thus the value of $\delta(c_1)$ is independent of the chosen lift. [/guided] [/step] [step:Prove additivity by lifting the sum with the sum of lifts] Let $c_1,d_1\in\ker\gamma$. Choose $b_1,e_1\in B_1$ such that \begin{align*} p_1(b_1)=c_1, \qquad p_1(e_1)=d_1. \end{align*} Let $a_0,f_0\in A_0$ be the unique elements satisfying \begin{align*} i_0(a_0)=\beta(b_1), \qquad i_0(f_0)=\beta(e_1). \end{align*} Since $\gamma$ is an $R$-module homomorphism, $\ker\gamma$ is a submodule of $C_1$, so $c_1+d_1\in\ker\gamma$. Also, \begin{align*} p_1(b_1+e_1)=p_1(b_1)+p_1(e_1)=c_1+d_1. \end{align*} Thus $b_1+e_1$ is a valid lift of $c_1+d_1$. Its associated element of $A_0$ is $a_0+f_0$, because \begin{align*} i_0(a_0+f_0)=i_0(a_0)+i_0(f_0)=\beta(b_1)+\beta(e_1)=\beta(b_1+e_1). \end{align*} Therefore \begin{align*} \delta(c_1+d_1) &=(a_0+f_0)+\operatorname{im}\alpha \\ &=(a_0+\operatorname{im}\alpha)+(f_0+\operatorname{im}\alpha) \\ &=\delta(c_1)+\delta(d_1). \end{align*} [/step] [step:Prove compatibility with scalar multiplication by lifting scalar multiples] Let $r\in R$ and $c_1\in\ker\gamma$. Choose $b_1\in B_1$ such that $p_1(b_1)=c_1$, and let $a_0\in A_0$ be the unique element satisfying \begin{align*} i_0(a_0)=\beta(b_1). \end{align*} Since $\ker\gamma$ is a submodule, $r c_1\in\ker\gamma$. Moreover \begin{align*} p_1(r b_1)=r p_1(b_1)=r c_1, \end{align*} so $r b_1$ is a valid lift of $r c_1$. The associated element of $A_0$ is $r a_0$, because \begin{align*} i_0(r a_0)=r i_0(a_0)=r\beta(b_1)=\beta(r b_1). \end{align*} Therefore \begin{align*} \delta(r c_1) &=r a_0+\operatorname{im}\alpha \\ &=r(a_0+\operatorname{im}\alpha) \\ &=r\delta(c_1). \end{align*} [/step] [step:Conclude that the connecting map is an $R$-module homomorphism] The previous steps show that $\delta(c_1)$ is independent of the chosen lift $b_1$, so $\delta:\ker\gamma\to\operatorname{coker}\alpha$ is a well-defined map. They also show that for all $c_1,d_1\in\ker\gamma$ and all $r\in R$, \begin{align*} \delta(c_1+d_1)=\delta(c_1)+\delta(d_1), \qquad \delta(r c_1)=r\delta(c_1). \end{align*} Hence $\delta$ is an $R$-module homomorphism. [/step]