[proofplan] We prove the equivalence by following the natural cycle of implications. The lifting definition of projectivity immediately gives sections of epimorphisms onto $P$, and such a section splits a free presentation of $P$, making $P$ a direct summand of a free module. A direct summand of a free module has the lifting property because maps out of a free module lift coordinatewise across surjections. Finally, the Hom formulation is exactly the same lifting property expressed as surjectivity of the postcomposition map on Hom groups. [/proofplan]

[step:Use projectivity to split every surjection onto $P$] Assume $P$ is projective, and let $q: M \to P$ be a surjective $R$-module homomorphism. Apply the lifting property in the definition of projectivity to the surjection $q: M \to P$ and to the homomorphism $\operatorname{id}_P: P \to P$. There exists an $R$-module homomorphism $s: P \to M$ such that \begin{align*} q \circ s = \operatorname{id}_P. \end{align*} Thus every surjection onto $P$ admits an $R$-linear section. [/step]

[step:Split a free presentation of $P$]Assume every surjective homomorphism onto $P$ admits an $R$-linear section. Choose a generating set $X \subset P$ for $P$ as a left $R$-module. Let $F$ be the free left $R$-module on $X$, with basis elements denoted by $e_x$ for $x \in X$. Define the $R$-module homomorphism \begin{align*} \pi: F &\to P \\ e_x &\mapsto x \end{align*} and extend $R$-linearly. Since $X$ generates $P$, the map $\pi$ is surjective. By the assumed splitting property, there exists an $R$-module homomorphism $s: P \to F$ such that \begin{align*} \pi \circ s = \operatorname{id}_P. \end{align*} Let $Q := \ker \pi$, considered as a left $R$-submodule of $F$. We prove that \begin{align*} F = s(P) \oplus Q. \end{align*} If $y \in s(P) \cap Q$, then $y = s(p)$ for some $p \in P$ and $\pi(y) = 0$. Hence \begin{align*} p = \operatorname{id}_P(p) = (\pi \circ s)(p) = \pi(y) = 0, \end{align*} so $y = s(0) = 0$. Thus $s(P) \cap Q = \{0\}$. For every $z \in F$, define \begin{align*} p_z &:= \pi(z) \in P, & q_z &:= z - s(\pi(z)) \in F. \end{align*} Then \begin{align*} \pi(q_z) &= \pi(z - s(\pi(z))) \\ &= \pi(z) - (\pi \circ s)(\pi(z)) \\ &= \pi(z) - \pi(z) \\ &= 0, \end{align*} so $q_z \in Q$, and \begin{align*} z = s(p_z) + q_z. \end{align*} Therefore $F = s(P) \oplus Q$. Since $s: P \to s(P)$ is an isomorphism with inverse $\pi|_{s(P)}: s(P) \to P$, the module $P$ is isomorphic to the direct summand $s(P)$ of the free module $F$.[/step]

[guided]We want to turn the abstract splitting condition into the concrete statement that $P$ sits inside a free module as one summand. First choose a generating set $X \subset P$. Let $F$ be the free left $R$-module with basis $\{e_x : x \in X\}$. The universal property of the free module gives an $R$-module homomorphism \begin{align*} \pi: F &\to P \\ e_x &\mapsto x, \end{align*} extended $R$-linearly. Because $X$ generates $P$, every element of $P$ is an $R$-linear combination of elements of $X$, so $\pi$ is surjective. By the assumed condition, the surjection $\pi: F \to P$ has an $R$-linear section. Thus there is an $R$-module homomorphism $s: P \to F$ satisfying \begin{align*} \pi \circ s = \operatorname{id}_P. \end{align*} This identity says that $s$ embeds $P$ into $F$ without losing information: if $s(p) = 0$, then applying $\pi$ gives $p = 0$. Now define $Q := \ker \pi$. We claim that $F = s(P) \oplus Q$. First the intersection is zero. If $y \in s(P) \cap Q$, then $y = s(p)$ for some $p \in P$, and also $\pi(y) = 0$ because $y \in Q$. Applying $\pi$ to $y = s(p)$ gives \begin{align*} p = \operatorname{id}_P(p) = (\pi \circ s)(p) = \pi(y) = 0, \end{align*} so $y = s(0) = 0$. Next every element of $F$ decomposes as a sum of one element from $s(P)$ and one element from $Q$. For $z \in F$, define \begin{align*} p_z &:= \pi(z) \in P, & q_z &:= z - s(\pi(z)) \in F. \end{align*} Then \begin{align*} \pi(q_z) &= \pi(z - s(\pi(z))) \\ &= \pi(z) - (\pi \circ s)(\pi(z)) \\ &= \pi(z) - \pi(z) \\ &= 0. \end{align*} Thus $q_z \in Q$, while $s(p_z) \in s(P)$, and \begin{align*} z = s(p_z) + q_z. \end{align*} So $F = s(P) \oplus Q$. Finally, $s: P \to s(P)$ is an isomorphism, with inverse $\pi|_{s(P)}: s(P) \to P$. Hence $P$ is isomorphic to a direct summand of the free module $F$.[/guided]

[step:Lift maps from a direct summand of a free module]Assume there exist a free left $R$-module $F$ and a left $R$-module $Q$ such that $F \cong P \oplus Q$. Replacing $F$ by an isomorphic free module if necessary, we may use $R$-module homomorphisms \begin{align*} i: P &\to F, & \rho: F &\to P \end{align*} such that $\rho \circ i = \operatorname{id}_P$. Let $q: M \to N$ be a surjective $R$-module homomorphism, and let $f: P \to N$ be an $R$-module homomorphism. We must construct an $R$-module homomorphism $\widetilde{f}: P \to M$ with $q \circ \widetilde{f} = f$. Let $(e_a)_{a \in A}$ be a basis of the free left $R$-module $F$. Define the $R$-module homomorphism \begin{align*} g: F &\to N \\ x &\mapsto f(\rho(x)). \end{align*} For each $a \in A$, the surjectivity of $q$ gives an element $m_a \in M$ such that \begin{align*} q(m_a) = g(e_a). \end{align*} By freeness of $F$, there is a unique $R$-module homomorphism \begin{align*} h: F &\to M \\ e_a &\mapsto m_a \end{align*} extended $R$-linearly. For every basis element $e_a$, we have \begin{align*} (q \circ h)(e_a) = q(m_a) = g(e_a). \end{align*} Since two $R$-module homomorphisms out of a free module are equal when they agree on a basis, $q \circ h = g$. Define \begin{align*} \widetilde{f}: P &\to M \\ p &\mapsto h(i(p)). \end{align*} Then \begin{align*} (q \circ \widetilde{f})(p) &= q(h(i(p))) \\ &= g(i(p)) \\ &= f(\rho(i(p))) \\ &= f(p) \end{align*} for every $p \in P$. Hence $q \circ \widetilde{f} = f$, so $P$ is projective.[/step]

[guided]The goal is to prove the lifting property for $P$. The direct summand hypothesis means that maps out of $P$ can be extended to maps out of a free module, and maps out of free modules are easy to lift because a free map is determined by the images of basis elements. Because $P$ is isomorphic to a direct summand of $F$, we may work with inclusion and projection maps \begin{align*} i: P &\to F, & \rho: F &\to P \end{align*} satisfying $\rho \circ i = \operatorname{id}_P$. Here $i$ identifies $P$ with its copy inside $F$, and $\rho$ projects $F$ back onto that summand. Let $q: M \to N$ be a surjective $R$-module homomorphism, and let $f: P \to N$ be an $R$-module homomorphism. We must find a map $\widetilde{f}: P \to M$ whose composition with $q$ is $f$. First extend $f$ from $P$ to $F$ by defining \begin{align*} g: F &\to N \\ x &\mapsto f(\rho(x)). \end{align*} This is an $R$-module homomorphism because it is the composition $f \circ \rho$. Now let $(e_a)_{a \in A}$ be a basis of the free module $F$. Since $q$ is surjective, each element $g(e_a) \in N$ has at least one preimage in $M$. Choose such an element $m_a \in M$ for each $a \in A$, so \begin{align*} q(m_a) = g(e_a). \end{align*} The defining property of the free module $F$ gives a unique $R$-module homomorphism \begin{align*} h: F &\to M \\ e_a &\mapsto m_a \end{align*} extended $R$-linearly. By construction, $q \circ h$ and $g$ agree on every basis element: \begin{align*} (q \circ h)(e_a) = q(m_a) = g(e_a). \end{align*} Since both maps are $R$-linear and $F$ is generated freely by the elements $e_a$, this equality on the basis implies $q \circ h = g$ on all of $F$. Finally restrict the lifted map $h$ back to the summand $P$ by defining \begin{align*} \widetilde{f}: P &\to M \\ p &\mapsto h(i(p)). \end{align*} For every $p \in P$, \begin{align*} (q \circ \widetilde{f})(p) &= q(h(i(p))) \\ &= g(i(p)) \\ &= f(\rho(i(p))) \\ &= f(p), \end{align*} because $\rho \circ i = \operatorname{id}_P$. Therefore $q \circ \widetilde{f} = f$, which is exactly the lifting property. Hence $P$ is projective.[/guided]

[step:Translate the lifting property into surjectivity of Hom maps] We prove that projectivity of $P$ is equivalent to condition 4. Assume first that $P$ is projective. Let $q: M \to N$ be a surjective $R$-module homomorphism. The induced map \begin{align*} q_*: \operatorname{Hom}_R(P, M) &\to \operatorname{Hom}_R(P, N) \\ \varphi &\mapsto q \circ \varphi \end{align*} is a homomorphism of abelian groups because postcomposition with the $R$-[linear map](/page/Linear%20Map) $q$ preserves addition of homomorphisms. To prove that $q_*$ is surjective, let $f \in \operatorname{Hom}_R(P, N)$. By projectivity of $P$, there exists $\widetilde{f} \in \operatorname{Hom}_R(P, M)$ such that \begin{align*} q \circ \widetilde{f} = f. \end{align*} Thus $q_*(\widetilde{f}) = f$, so $q_*$ is surjective. Conversely, assume condition 4. Let $q: M \to N$ be a surjective $R$-module homomorphism, and let $f: P \to N$ be an $R$-module homomorphism. By condition 4, the induced map \begin{align*} q_*: \operatorname{Hom}_R(P, M) &\to \operatorname{Hom}_R(P, N) \\ \varphi &\mapsto q \circ \varphi \end{align*} is surjective. Since $f \in \operatorname{Hom}_R(P, N)$, there exists $\widetilde{f} \in \operatorname{Hom}_R(P, M)$ such that \begin{align*} q_*(\widetilde{f}) = f. \end{align*} By the definition of $q_*$, this says $q \circ \widetilde{f} = f$. Therefore $P$ satisfies the lifting property, so $P$ is projective. [/step]

[step:Conclude all four conditions are equivalent] The first step proves $(1) \implies (2)$, the second proves $(2) \implies (3)$, and the third proves $(3) \implies (1)$. Hence conditions 1, 2, and 3 are equivalent. The fourth step proves $(1) \iff (4)$. Therefore all four stated conditions are equivalent. [/step]