[proofplan] The first two conditions are the same extension property, with condition (2) phrased only for the first map in a short exact sequence. Injectivity gives a retraction from any module containing $I$, and such a retraction splits every short exact sequence starting with $I$. Conversely, if every embedding of $I$ admits a retraction, then the extension property follows by constructing the pushout of a map $A \to B$ along a map $A \to I$ and retracting the induced copy of $I$ inside that pushout. [/proofplan]

[step:Identify injectivity with extension across the first map of a short exact sequence] Condition (1) says precisely that every left $R$-module homomorphism $f: A \to I$ extends across every injective left $R$-module homomorphism $\alpha: A \to B$. If \begin{align*} 0 \longrightarrow A \xrightarrow{\iota} B \xrightarrow{\pi} C \longrightarrow 0 \end{align*} is short exact, then $\iota: A \to B$ is injective, so condition (1) gives condition (2). Conversely, if $\alpha: A \to B$ is an injective left $R$-module homomorphism, then the quotient map $q: B \to B/\alpha(A)$ gives a short exact sequence \begin{align*} 0 \longrightarrow A \xrightarrow{\alpha} B \xrightarrow{q} B/\alpha(A) \longrightarrow 0, \end{align*} where $A$ is identified with its image through the injective map $\alpha$. Applying condition (2) to this short exact sequence gives the extension required in condition (1). Hence conditions (1) and (2) are equivalent. [/step]

[step:Use injectivity to split every short exact sequence starting with $I$]Assume condition (1). Let \begin{align*} 0 \longrightarrow I \xrightarrow{u} M \xrightarrow{v} N \longrightarrow 0 \end{align*} be a short exact sequence of left $R$-modules. Since $u: I \to M$ is injective and $\operatorname{id}_I: I \to I$ is a left $R$-module homomorphism, injectivity of $I$ gives a left $R$-module homomorphism \begin{align*} r: M \to I \end{align*} such that $r \circ u = \operatorname{id}_I$. We construct a splitting map for $v$. For $n \in N$, choose $m \in M$ with $v(m) = n$, possible because $v$ is surjective. Define \begin{align*} s: N &\to M \\ n &\mapsto m - u(r(m)). \end{align*} This definition is independent of the choice of $m$. Indeed, if $m' \in M$ also satisfies $v(m') = n$, then $v(m - m') = 0$, so exactness gives $m - m' \in \operatorname{im} u$. Thus there exists $i \in I$ such that $m - m' = u(i)$. Since $r \circ u = \operatorname{id}_I$, \begin{align*} m - u(r(m)) - \bigl(m' - u(r(m'))\bigr) &= m - m' - u(r(m) - r(m')) \\ &= u(i) - u(r(m - m')) \\ &= u(i) - u(r(u(i))) \\ &= u(i) - u(i) \\ &= 0. \end{align*} The map $s$ is $R$-linear because it is induced by the $R$-linear operation $m \mapsto m - u(r(m))$ on representatives, and it satisfies \begin{align*} (v \circ s)(n) &= v(m - u(r(m))) \\ &= v(m) - v(u(r(m))) \\ &= n - 0 \\ &= n. \end{align*} Therefore $v \circ s = \operatorname{id}_N$, so the short exact sequence splits. Hence condition (1) implies condition (3).[/step]

[guided]Assume condition (1), so $I$ has the extension property across every injective left $R$-module homomorphism. Let \begin{align*} 0 \longrightarrow I \xrightarrow{u} M \xrightarrow{v} N \longrightarrow 0 \end{align*} be a short exact sequence. Exactness at $I$ means $u: I \to M$ is injective. We apply injectivity of $I$ to the injective map $u: I \to M$ and to the homomorphism $\operatorname{id}_I: I \to I$. This gives a left $R$-module homomorphism \begin{align*} r: M \to I \end{align*} such that \begin{align*} r \circ u = \operatorname{id}_I. \end{align*} Thus $r$ is a retraction of the inclusion $u$. To show that the original sequence splits, we produce a right inverse to $v$. Let $n \in N$. Since $v$ is surjective, choose $m \in M$ with $v(m) = n$. The expression $m - u(r(m))$ removes from $m$ its component lying in the copy of $I$. Define \begin{align*} s: N &\to M \\ n &\mapsto m - u(r(m)). \end{align*} We must verify that this does not depend on the chosen representative $m$. Suppose $m' \in M$ also satisfies $v(m') = n$. Then $v(m - m') = 0$, so $m - m' \in \ker v$. Exactness gives $\ker v = \operatorname{im} u$, hence there exists $i \in I$ with $m - m' = u(i)$. Using $R$-linearity of $r$ and $r \circ u = \operatorname{id}_I$, \begin{align*} m - u(r(m)) - \bigl(m' - u(r(m'))\bigr) &= m - m' - u(r(m) - r(m')) \\ &= u(i) - u(r(m - m')) \\ &= u(i) - u(r(u(i))) \\ &= u(i) - u(i) \\ &= 0. \end{align*} So $s$ is well-defined. The same representative calculation shows that $s$ is $R$-linear. Finally, \begin{align*} (v \circ s)(n) &= v(m - u(r(m))) \\ &= v(m) - v(u(r(m))) \\ &= n - 0 \\ &= n, \end{align*} because $\operatorname{im} u = \ker v$. Thus $v \circ s = \operatorname{id}_N$, so the sequence splits.[/guided]

[step:Specialize split exact sequences to embeddings of $I$] Assume condition (3). Let $M$ be a left $R$-module, and let $j: I \to M$ be an injective left $R$-module homomorphism. Let $q: M \to M/j(I)$ be the quotient homomorphism. Then \begin{align*} 0 \longrightarrow I \xrightarrow{j} M \xrightarrow{q} M/j(I) \longrightarrow 0 \end{align*} is a short exact sequence. By condition (3), this sequence splits. In particular, there exists a left $R$-module homomorphism \begin{align*} r: M \to I \end{align*} such that $r \circ j = \operatorname{id}_I$. Hence condition (4) holds. [/step]

[step:Build a pushout to recover the extension property]Assume condition (4). Let $\alpha: A \to B$ be an injective left $R$-module homomorphism, and let $f: A \to I$ be a left $R$-module homomorphism. We must construct a left $R$-module homomorphism $g: B \to I$ such that $g \circ \alpha = f$. Define the left $R$-module $P$ by \begin{align*} P := (B \oplus I)/K, \end{align*} where $K \subset B \oplus I$ is the submodule \begin{align*} K := \{(\alpha(a), -f(a)) : a \in A\}. \end{align*} Define left $R$-module homomorphisms \begin{align*} j_B: B &\to P, & b &\mapsto [(b,0)], \\ j_I: I &\to P, & i &\mapsto [(0,i)], \end{align*} where $[(b,i)]$ denotes the class of $(b,i) \in B \oplus I$ modulo $K$. The map $j_I$ is injective. If $j_I(i) = 0$, then $(0,i) \in K$, so there exists $a \in A$ such that \begin{align*} (0,i) = (\alpha(a), -f(a)). \end{align*} Thus $\alpha(a) = 0$. Since $\alpha$ is injective, $a = 0$, and then $i = -f(0) = 0$. Hence $j_I$ embeds $I$ as a submodule of $P$. By condition (4), applied to the embedding $j_I: I \to P$, there exists a left $R$-module homomorphism \begin{align*} \rho: P \to I \end{align*} such that $\rho \circ j_I = \operatorname{id}_I$. Define \begin{align*} g: B &\to I \\ b &\mapsto \rho(j_B(b)). \end{align*} Then $g$ is a left $R$-module homomorphism. For every $a \in A$, the definition of $K$ gives \begin{align*} j_B(\alpha(a)) &= [(\alpha(a),0)] \\ &= [(0,f(a))] \\ &= j_I(f(a)). \end{align*} Therefore \begin{align*} (g \circ \alpha)(a) &= \rho(j_B(\alpha(a))) \\ &= \rho(j_I(f(a))) \\ &= f(a). \end{align*} Thus $g \circ \alpha = f$, so $I$ is injective. Hence condition (4) implies condition (1).[/step]

[guided]Assume condition (4): every embedding of $I$ into a left $R$-module admits a retraction. We want to prove injectivity of $I$. Therefore we start with an arbitrary injective homomorphism $\alpha: A \to B$ and an arbitrary homomorphism $f: A \to I$, and we must extend $f$ to a homomorphism $B \to I$. The obstruction is that $f$ is defined only on $A$, while the desired extension is defined on $B$. The pushout forces the two maps out of $A$ to agree in a single module. Define \begin{align*} P := (B \oplus I)/K, \end{align*} where \begin{align*} K := \{(\alpha(a), -f(a)) : a \in A\}. \end{align*} This is a submodule of $B \oplus I$ because $\alpha$ and $f$ are $R$-linear. Let $[(b,i)]$ denote the class of $(b,i) \in B \oplus I$ modulo $K$. Define \begin{align*} j_B: B &\to P, & b &\mapsto [(b,0)], \\ j_I: I &\to P, & i &\mapsto [(0,i)]. \end{align*} Both maps are left $R$-module homomorphisms because they are induced from the coordinate inclusions into the direct sum followed by the quotient map. We need to apply condition (4), so we must check that $j_I$ is injective. Suppose $j_I(i) = 0$. Then $(0,i) \in K$, so there exists $a \in A$ such that \begin{align*} (0,i) = (\alpha(a), -f(a)). \end{align*} Comparing the first coordinate gives $\alpha(a) = 0$. Since $\alpha$ is injective, $a = 0$. Comparing the second coordinate then gives $i = -f(0) = 0$. Thus $j_I$ is injective. Condition (4) now applies to the embedding $j_I: I \to P$. Hence there exists a left $R$-module homomorphism \begin{align*} \rho: P \to I \end{align*} such that \begin{align*} \rho \circ j_I = \operatorname{id}_I. \end{align*} Define \begin{align*} g: B &\to I \\ b &\mapsto \rho(j_B(b)). \end{align*} This map is $R$-linear as a composition of $R$-linear maps. It remains to verify that $g$ extends $f$. For $a \in A$, the element $(\alpha(a),-f(a))$ lies in $K$, so it becomes zero in the quotient $P$. Equivalently, \begin{align*} [(\alpha(a),0)] = [(0,f(a))]. \end{align*} Thus \begin{align*} j_B(\alpha(a)) = j_I(f(a)). \end{align*} Applying $\rho$ gives \begin{align*} (g \circ \alpha)(a) &= \rho(j_B(\alpha(a))) \\ &= \rho(j_I(f(a))) \\ &= f(a), \end{align*} because $\rho \circ j_I = \operatorname{id}_I$. Since this holds for every $a \in A$, we have $g \circ \alpha = f$. This is exactly the injectivity extension property for $I$.[/guided]

[step:Combine the implications] We have proved condition (1) is equivalent to condition (2), condition (1) implies condition (3), condition (3) implies condition (4), and condition (4) implies condition (1). Therefore all four conditions are equivalent. [/step]