[proofplan] We prove that a compatible direct-sum decomposition, a right inverse to $p$, and a left inverse to $i$ determine one another. From a splitting isomorphism, the section and retraction are obtained by composing with the standard inclusion and projection maps of $A \oplus C$. Conversely, a section builds an explicit isomorphism $A \oplus C \to B$ by $(a,c) \mapsto i(a)+s(c)$. A retraction gives the complementary submodule $\ker r$, and exactness shows that $p|_{\ker r}: \ker r \to C$ is an isomorphism, which again gives a compatible direct-sum decomposition. [/proofplan] [step:Extract a section and a retraction from a compatible splitting] Assume the sequence is split. Let $\Phi: A \oplus C \to B$ be an $R$-module isomorphism satisfying \begin{align*} \Phi(a,0) &= i(a), \\ p(\Phi(a,c)) &= c \end{align*} for every $(a,c) \in A \oplus C$. Define the standard inclusion of the second summand \begin{align*} \jmath_C: C &\to A \oplus C \\ c &\mapsto (0,c), \end{align*} and define \begin{align*} s: C &\to B \\ c &\mapsto \Phi(0,c). \end{align*} Then $s=\Phi \circ \jmath_C$, so $s$ is an $R$-module homomorphism. For every $c \in C$, \begin{align*} (p \circ s)(c)=p(\Phi(0,c))=c, \end{align*} so $p \circ s=\operatorname{id}_C$. Now define \begin{align*} \pi_A: A \oplus C &\to A \\ (a,c) &\mapsto a \end{align*} and \begin{align*} r: B &\to A \\ b &\mapsto \pi_A(\Phi^{-1}(b)). \end{align*} Since $\Phi^{-1}$ and $\pi_A$ are $R$-module homomorphisms, $r$ is an $R$-module homomorphism. 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 $r \circ i=\operatorname{id}_A$. [/step] [step:Build a compatible splitting from a section] Assume there exists an $R$-module homomorphism $s: C \to B$ such that $p \circ s=\operatorname{id}_C$. Define \begin{align*} \Phi: A \oplus C &\to B \\ (a,c) &\mapsto i(a)+s(c). \end{align*} Because $i$ and $s$ are $R$-module homomorphisms, $\Phi$ is an $R$-module homomorphism. We prove that $\Phi$ is surjective. Let $b \in B$. Since $p \circ s=\operatorname{id}_C$, \begin{align*} p(b-s(p(b))) &= p(b)-p(s(p(b))) \\ &= p(b)-p(b) \\ &= 0. \end{align*} Hence $b-s(p(b)) \in \ker p$. By exactness, $\ker p=\operatorname{im} i$, so there exists $a \in A$ such that \begin{align*} i(a)=b-s(p(b)). \end{align*} Therefore \begin{align*} \Phi(a,p(b)) &= i(a)+s(p(b)) \\ &= b. \end{align*} Thus $\Phi$ is surjective. We prove that $\Phi$ is injective. Suppose $(a,c) \in A \oplus C$ satisfies $\Phi(a,c)=0$. Applying $p$ gives \begin{align*} 0 &= p(\Phi(a,c)) \\ &= p(i(a)+s(c)) \\ &= p(i(a))+p(s(c)) \\ &= 0+c \\ &= c, \end{align*} where $p(i(a))=0$ because $\operatorname{im} i \subset \ker p$. Hence $c=0$. Then \begin{align*} 0=\Phi(a,0)=i(a). \end{align*} Since $i$ is injective, $a=0$. Thus $\ker \Phi=\{(0,0)\}$, so $\Phi$ is injective. Therefore $\Phi$ is an $R$-module isomorphism. Moreover, for every $a \in A$ and every $(a,c) \in A \oplus C$, \begin{align*} \Phi(a,0)&=i(a), \\ p(\Phi(a,c)) &=p(i(a)+s(c)) \\ &=0+c \\ &=c. \end{align*} Hence the short exact sequence is split. [/step] [step:Identify the kernel of a retraction as a complementary copy of $C$] Assume there exists an $R$-module homomorphism $r: B \to A$ such that $r \circ i=\operatorname{id}_A$. Define the submodule \begin{align*} K:=\ker r \subset B. \end{align*} Let \begin{align*} q: K &\to C \\ b &\mapsto p(b) \end{align*} be the restriction of $p$ to $K$. Since $p$ is an $R$-module homomorphism and $K$ is a submodule of $B$, $q$ is an $R$-module homomorphism. We first prove that $q$ is injective. Let $b \in K$ and suppose $q(b)=0$. Then $p(b)=0$, so $b \in \ker p$. By exactness, $\ker p=\operatorname{im} i$, so there exists $a \in A$ such that $b=i(a)$. Since $b \in K=\ker r$, \begin{align*} 0 &=r(b) \\ &=r(i(a)) \\ &=a. \end{align*} Thus $b=i(0)=0$, and $q$ is injective. We next prove that $q$ is surjective. Let $c \in C$. Since $p$ is surjective, there exists $b \in B$ such that $p(b)=c$. Define \begin{align*} b_0:=b-i(r(b)) \in B. \end{align*} Then \begin{align*} r(b_0) &=r(b)-r(i(r(b))) \\ &=r(b)-r(b) \\ &=0, \end{align*} so $b_0 \in K$. Also, \begin{align*} q(b_0) &=p(b-i(r(b))) \\ &=p(b)-p(i(r(b))) \\ &=c-0 \\ &=c, \end{align*} because $\operatorname{im} i \subset \ker p$. Hence $q$ is surjective. Thus $q: K \to C$ is an $R$-module isomorphism. Let \begin{align*} t: C &\to K \end{align*} denote its inverse, and let \begin{align*} \ell: K &\to B \\ b &\mapsto b \end{align*} be the inclusion map. Define \begin{align*} s: C &\to B \\ c &\mapsto \ell(t(c)). \end{align*} Then $s$ is an $R$-module homomorphism, and for every $c \in C$, \begin{align*} (p \circ s)(c) &=p(\ell(t(c))) \\ &=q(t(c)) \\ &=c. \end{align*} So $p \circ s=\operatorname{id}_C$. [/step] [step:Conclude the equivalence of the three criteria] The first step proves that condition 1 implies both condition 2 and condition 3. The second step proves that condition 2 implies condition 1. The third step proves that condition 3 implies condition 2, and the second step then gives condition 1. Therefore conditions 1, 2, and 3 are equivalent. [/step]