[proofplan] We prove that each of the three formulations gives a direct-sum decomposition of $B$ into a copy of $A$ and a complementary copy of $C$. A section $s:C\to B$ produces an explicit isomorphism $A\oplus C\to B$ by $(a,c)\mapsto i(a)+s(c)$. A compatible direct-sum isomorphism immediately gives both a section and a retraction by composing with the standard inclusion and projection maps. Finally, a retraction $r:B\to A$ gives the complementary submodule $\ker r$, and exactness shows that $p$ restricts to an isomorphism $\ker r\to C$. [/proofplan] [step:Construct a compatible direct sum decomposition from a section of $p$] Assume there exists an $R$-[linear map](/page/Linear%20Map) $s:C\to B$ such that $p\circ s=\operatorname{id}_C$. Define the $R$-linear map \begin{align*} \Phi:A\oplus C&\to B\\ (a,c)&\mapsto i(a)+s(c). \end{align*} We prove that $\Phi$ is an isomorphism and satisfies the compatibility conditions. First, $\Phi(a,0)=i(a)+s(0)=i(a)$ for every $a\in A$, since $s$ is $R$-linear. Also, for every $a\in A$ and $c\in C$, \begin{align*} p(\Phi(a,c)) &=p(i(a)+s(c))\\ &=p(i(a))+p(s(c))\\ &=0+c\\ &=c, \end{align*} because $\operatorname{im} i=\ker p$ implies $p\circ i=0$, and $p\circ s=\operatorname{id}_C$. To prove injectivity, let $(a,c)\in A\oplus C$ satisfy $\Phi(a,c)=0$. Applying $p$ gives \begin{align*} 0=p(0)=p(\Phi(a,c))=c. \end{align*} Thus $c=0$, and then $0=\Phi(a,0)=i(a)$. Since $i$ is injective, $a=0$. Hence $\ker\Phi=\{(0,0)\}$. To prove surjectivity, let $b\in B$. Define $c:=p(b)\in C$. Then \begin{align*} p(b-s(c)) &=p(b)-p(s(c))\\ &=c-c\\ &=0. \end{align*} Thus $b-s(c)\in\ker p=\operatorname{im} i$. Hence there exists $a\in A$ such that $i(a)=b-s(c)$. Therefore \begin{align*} b=i(a)+s(c)=\Phi(a,c). \end{align*} So $\Phi$ is surjective. Therefore $\Phi:A\oplus C\to B$ is an $R$-module isomorphism compatible with $i$ and $p$. [guided] Assume we are given an $R$-linear section \begin{align*} s:C\to B \end{align*} with $p\circ s=\operatorname{id}_C$. The section chooses, for each $c\in C$, a representative $s(c)\in B$ mapping back to $c$. The natural candidate for the direct-sum isomorphism is therefore \begin{align*} \Phi:A\oplus C&\to B\\ (a,c)&\mapsto i(a)+s(c). \end{align*} This map is $R$-linear because $i$ and $s$ are $R$-linear and addition in $B$ is $R$-bilinear. We first verify compatibility with the sequence. For $a\in A$, \begin{align*} \Phi(a,0)=i(a)+s(0)=i(a), \end{align*} because every $R$-linear map sends $0$ to $0$. For $a\in A$ and $c\in C$, \begin{align*} p(\Phi(a,c)) &=p(i(a)+s(c))\\ &=p(i(a))+p(s(c))\\ &=0+c\\ &=c. \end{align*} Here $p(i(a))=0$ because $\operatorname{im} i=\ker p$, and $p(s(c))=c$ because $p\circ s=\operatorname{id}_C$. Now we prove injectivity. Suppose $\Phi(a,c)=0$ for some $(a,c)\in A\oplus C$. Applying $p$ removes the $A$-part and keeps the $C$-part: \begin{align*} 0=p(0)=p(\Phi(a,c))=c. \end{align*} Thus $c=0$. Substituting this back gives \begin{align*} 0=\Phi(a,0)=i(a). \end{align*} Since exactness gives that $i$ is injective, $a=0$. Therefore the only element in $\ker\Phi$ is $(0,0)$, so $\Phi$ is injective. Finally we prove surjectivity. Let $b\in B$. The element $p(b)\in C$ is the $C$-component we must assign to $b$, so define $c:=p(b)$. Subtracting the chosen lift $s(c)$ kills the image under $p$: \begin{align*} p(b-s(c)) &=p(b)-p(s(c))\\ &=c-c\\ &=0. \end{align*} Hence $b-s(c)\in\ker p$. Exactness says $\ker p=\operatorname{im} i$, so there exists $a\in A$ with $i(a)=b-s(c)$. Thus \begin{align*} b=i(a)+s(c)=\Phi(a,c). \end{align*} Every $b\in B$ lies in the image of $\Phi$, so $\Phi$ is surjective. Therefore $\Phi$ is a compatible $R$-module isomorphism $A\oplus C\to B$. [/guided] [/step] [step:Recover a section and a retraction from a compatible direct sum decomposition] Assume there exists an $R$-module isomorphism $\Phi:A\oplus C\to B$ satisfying \begin{align*} \Phi(a,0)&=i(a),\\ p(\Phi(a,c))&=c. \end{align*} Define the $R$-linear map \begin{align*} s:C&\to B\\ c&\mapsto \Phi(0,c). \end{align*} Then for every $c\in C$, \begin{align*} (p\circ s)(c)=p(\Phi(0,c))=c, \end{align*} so $p\circ s=\operatorname{id}_C$. Let $\pi_A:A\oplus C\to A$ be the standard projection, defined by $\pi_A(a,c)=a$. Define the $R$-linear map \begin{align*} r:B&\to A\\ b&\mapsto \pi_A(\Phi^{-1}(b)). \end{align*} For every $a\in A$, \begin{align*} (r\circ i)(a) &=\pi_A(\Phi^{-1}(i(a)))\\ &=\pi_A(\Phi^{-1}(\Phi(a,0)))\\ &=\pi_A(a,0)\\ &=a. \end{align*} Thus $r\circ i=\operatorname{id}_A$. Hence the compatible direct-sum decomposition implies both the existence of a section of $p$ and the existence of a retraction of $i$. [/step] [step:Construct a section of $p$ from a retraction of $i$] Assume there exists an $R$-linear map $r:B\to A$ such that $r\circ i=\operatorname{id}_A$. Let $K:=\ker r\subset B$. We show that the restriction of $p$ to $K$ is an $R$-module isomorphism \begin{align*} p|_K:K\to C. \end{align*} First, $p|_K$ is injective. Let $k\in K$ satisfy $p(k)=0$. Then $k\in\ker p=\operatorname{im} i$, so there exists $a\in A$ such that $k=i(a)$. Since $k\in K=\ker r$, \begin{align*} 0=r(k)=r(i(a))=a. \end{align*} Therefore $k=i(0)=0$. Second, $p|_K$ is surjective. Let $c\in C$. Since $p:B\to C$ is surjective, choose $b\in B$ such that $p(b)=c$. Define \begin{align*} k:=b-i(r(b))\in B. \end{align*} Then \begin{align*} r(k) &=r(b)-r(i(r(b)))\\ &=r(b)-r(b)\\ &=0, \end{align*} so $k\in K$. Also, \begin{align*} p(k) &=p(b)-p(i(r(b)))\\ &=c-0\\ &=c, \end{align*} because $\operatorname{im} i=\ker p$ implies $p\circ i=0$. Hence $p|_K$ is surjective. Thus $p|_K:K\to C$ is an $R$-module isomorphism. Define \begin{align*} s:C&\to B\\ c&\mapsto (p|_K)^{-1}(c). \end{align*} This map is $R$-linear because $(p|_K)^{-1}:C\to K$ is an $R$-module isomorphism and the inclusion $K\subset B$ is $R$-linear. For every $c\in C$, \begin{align*} (p\circ s)(c)=p((p|_K)^{-1}(c))=c. \end{align*} Therefore $p\circ s=\operatorname{id}_C$. [guided] Assume we are given an $R$-linear retraction \begin{align*} r:B\to A \end{align*} with $r\circ i=\operatorname{id}_A$. The retraction identifies the $A$-part of an element of $B$. The complementary part should therefore be the part killed by $r$, so define \begin{align*} K:=\ker r\subset B. \end{align*} Since $r$ is $R$-linear, $K$ is an $R$-submodule of $B$. We prove that $p$ maps $K$ isomorphically onto $C$. First consider injectivity of the restricted map \begin{align*} p|_K:K\to C. \end{align*} Let $k\in K$ and suppose $p(k)=0$. Then $k\in\ker p$. Exactness gives $\ker p=\operatorname{im} i$, so there exists $a\in A$ such that $k=i(a)$. But $k\in K=\ker r$, hence \begin{align*} 0=r(k)=r(i(a))=(r\circ i)(a)=a. \end{align*} Therefore $k=i(a)=i(0)=0$. Thus $p|_K$ is injective. Now we prove surjectivity. Let $c\in C$. Since $p:B\to C$ is surjective, choose $b\in B$ with $p(b)=c$. The element $r(b)\in A$ is the $A$-component of $b$, so subtract its image under $i$: \begin{align*} k:=b-i(r(b)). \end{align*} We check that $k\in K$. Using $R$-linearity of $r$ and the identity $r\circ i=\operatorname{id}_A$, \begin{align*} r(k) &=r(b)-r(i(r(b)))\\ &=r(b)-r(b)\\ &=0. \end{align*} Thus $k\in\ker r=K$. We also check that $k$ maps to $c$ under $p$: \begin{align*} p(k) &=p(b)-p(i(r(b)))\\ &=c-0\\ &=c. \end{align*} The equality $p(i(r(b)))=0$ follows from $\operatorname{im} i=\ker p$. Therefore every $c\in C$ is hit by $p|_K$, so $p|_K$ is surjective. We have shown that $p|_K:K\to C$ is an $R$-module isomorphism. Its inverse lets us choose the unique element of $K$ lying above each $c\in C$. Define \begin{align*} s:C&\to B\\ c&\mapsto (p|_K)^{-1}(c), \end{align*} where we regard $(p|_K)^{-1}(c)\in K$ as an element of $B$ through the inclusion $K\subset B$. Since $p|_K$ is an $R$-module isomorphism, its inverse is $R$-linear, and hence $s$ is $R$-linear. Finally, \begin{align*} (p\circ s)(c)=p((p|_K)^{-1}(c))=c \end{align*} for every $c\in C$. Therefore $s$ is a section of $p$. [/guided] [/step] [step:Conclude the equivalence of the three splitting criteria] The first step proves that condition 1 implies condition 3. The second step proves that condition 3 implies both condition 1 and condition 2. The third step proves that condition 2 implies condition 1. Hence each of conditions 1, 2, and 3 implies the other two, so the three conditions are equivalent. This completes the proof. [/step]