[proofplan] We prove the equivalence by establishing the cycle of implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. For $(1) \Rightarrow (2)$, we present $P$ as a quotient of a [free module](/page/Free%20Module) and use the lifting property to split the surjection, making $P$ a [direct summand](/page/Direct%20Sum). For $(2) \Rightarrow (3)$, we show that $\operatorname{Hom}_R(F, -)$ is [exact](/page/Exact%20Sequence) for any free module $F$, then use the direct-sum decomposition $F \cong P \oplus Q$ to extract exactness for $\operatorname{Hom}_R(P, -)$. For $(3) \Rightarrow (1)$, right-exactness of $\operatorname{Hom}_R(P, -)$ applied to an arbitrary surjection immediately yields the lifting property. [/proofplan]

[step:Present $P$ as a quotient of a free module and lift the identity to split the surjection — $(1) \Rightarrow (2)$]Let $\{p_i\}_{i \in I}$ be a generating set for $P$ (for instance, $I = P$ and $p_i = i$ for each $i \in P$). Define the free $R$-module $F := R^{(I)} = \bigoplus_{i \in I} R$, and define the $R$-module homomorphism \begin{align*} \sigma: F &\to P \\ (r_i)_{i \in I} &\mapsto \sum_{i \in I} r_i p_i, \end{align*} where the sum is finite since elements of $F$ have only finitely many nonzero components. The map $\sigma$ is surjective because $\{p_i\}_{i \in I}$ generates $P$. Since $P$ is [projective](/theorems/???) and $\sigma: F \to P$ is surjective, applying the lifting property to the identity map $\operatorname{id}_P: P \to P$ yields an $R$-module homomorphism $s: P \to F$ satisfying $\sigma \circ s = \operatorname{id}_P$. The map $s$ is therefore a section of $\sigma$. Define $Q := \ker(\sigma)$. We verify that the internal [direct-sum](/page/Direct%20Sum) decomposition $F = s(P) \oplus Q$ holds. First, if $x \in s(P) \cap Q$, then $x = s(p)$ for some $p \in P$ and $\sigma(x) = 0$; but $\sigma(s(p)) = p$, so $p = 0$ and $x = s(0) = 0$. Second, for any $x \in F$, write $x = s(\sigma(x)) + (x - s(\sigma(x)))$. The first summand lies in $s(P)$, and the second lies in $Q$ since $\sigma(x - s(\sigma(x))) = \sigma(x) - \sigma(s(\sigma(x))) = \sigma(x) - \sigma(x) = 0$. Thus $F = s(P) \oplus Q$. The section $s: P \to s(P)$ is an isomorphism (with inverse $\sigma|_{s(P)}$), so $F \cong P \oplus Q$.[/step]

[guided]We need to exhibit a [free module](/page/Free%20Module) containing $P$ as a [direct summand](/page/Direct%20Sum). The strategy is: (a) present $P$ as a quotient of a free module, then (b) use projectivity to split the quotient map. **Constructing the free module.** Let $\{p_i\}_{i \in I}$ be a generating set for $P$. Any generating set works; one can take $I = P$ itself. Define the free $R$-module $F := R^{(I)} = \bigoplus_{i \in I} R$ with standard basis $\{e_i\}_{i \in I}$, and define the surjection \begin{align*} \sigma: F &\to P \\ (r_i)_{i \in I} &\mapsto \sum_{i \in I} r_i p_i. \end{align*} This is a well-defined [R-module homomorphism](/page/Module%20Homomorphism) because elements of $F$ have only finitely many nonzero components, so the sum is finite. It is surjective because $\{p_i\}_{i \in I}$ generates $P$: every element of $P$ is a finite $R$-linear combination of the $p_i$. **Splitting the surjection.** We now use the projective lifting property. Consider the diagram \begin{align*} &P \\ &\downarrow \operatorname{id}_P \\ F \xrightarrow{\;\sigma\;} &P \to 0. \end{align*} Since $\sigma$ is surjective and $P$ is [projective](/theorems/???), there exists an $R$-module homomorphism $s: P \to F$ with $\sigma \circ s = \operatorname{id}_P$. This map $s$ is a section: it "lifts" each element of $P$ back into $F$. **Decomposing $F$.** Why does the existence of a section give a direct-sum decomposition? Define $Q := \ker(\sigma)$. We claim $F = s(P) \oplus Q$. *Intersection is trivial:* Suppose $x \in s(P) \cap Q$. Then $x = s(p)$ for some $p \in P$ and $\sigma(x) = 0$. But $\sigma(s(p)) = (\sigma \circ s)(p) = \operatorname{id}_P(p) = p$, so $p = 0$, hence $x = s(0) = 0$. *Together they span $F$:* For any $x \in F$, write $x = s(\sigma(x)) + (x - s(\sigma(x)))$. The first summand lies in $s(P)$. The second lies in $Q$ because $\sigma(x - s(\sigma(x))) = \sigma(x) - \sigma(s(\sigma(x))) = \sigma(x) - \sigma(x) = 0$. Since $s$ is injective ($\sigma \circ s = \operatorname{id}_P$ forces injectivity: if $s(p) = s(p')$ then $p = \sigma(s(p)) = \sigma(s(p')) = p'$), the restriction $s: P \to s(P)$ is an $R$-module isomorphism with inverse $\sigma|_{s(P)}$. Therefore $F \cong P \oplus Q$, where $F$ is free.[/guided]

[step:Reduce exactness for $\operatorname{Hom}_R(P, -)$ to exactness for $\operatorname{Hom}_R(F, -)$ via the direct-sum decomposition — $(2) \Rightarrow (3)$]Let $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$ be a [short exact sequence](/page/Exact%20Sequence) of $R$-modules. By hypothesis, there exist a [free](/page/Free%20Module) $R$-module $F$ and an $R$-module $Q$ with $F \cong P \oplus Q$. [claim:$\operatorname{Hom}_R(F, -)$ is exact for any free module $F$] Let $F = \bigoplus_{i \in I} R$ be a free module with basis $\{e_i\}_{i \in I}$. For any $R$-module $M$, the map \begin{align*} \Phi_M: \operatorname{Hom}_R(F, M) &\to \prod_{i \in I} M \\ \varphi &\mapsto (\varphi(e_i))_{i \in I} \end{align*} is an isomorphism of abelian groups: an [R-module homomorphism](/page/Module%20Homomorphism) out of $F$ is determined by its values on the basis, and any assignment of basis elements to elements of $M$ extends uniquely to an $R$-module homomorphism. This identification is natural in $M$: for any $R$-module homomorphism $f: M \to N$, the diagram \begin{align*} \operatorname{Hom}_R(F, M) &\xrightarrow{\Phi_M} \prod_{i \in I} M \\ \downarrow f_* \quad &\qquad \downarrow \prod f \\ \operatorname{Hom}_R(F, N) &\xrightarrow{\Phi_N} \prod_{i \in I} N \end{align*} commutes, where $f_*(\varphi) = f \circ \varphi$ and $(\prod f)((m_i)) = (f(m_i))$. Now consider the short exact sequence $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$. Applying the natural isomorphism $\Phi$ converts the sequence \begin{align*} 0 \to \operatorname{Hom}_R(F, A) \xrightarrow{\iota_*} \operatorname{Hom}_R(F, B) \xrightarrow{\pi_*} \operatorname{Hom}_R(F, C) \to 0 \end{align*} into the sequence \begin{align*} 0 \to \prod_{i \in I} A \xrightarrow{\prod \iota} \prod_{i \in I} B \xrightarrow{\prod \pi} \prod_{i \in I} C \to 0. \end{align*} The latter is exact because the maps $\prod \iota$ and $\prod \pi$ act componentwise: $(\prod \iota)((a_i)) = (\iota(a_i))$ and $(\prod \pi)((b_i)) = (\pi(b_i))$. Therefore $(b_i) \in \ker(\prod \pi)$ if and only if $\pi(b_i) = 0$ for each $i$, i.e., $b_i \in \ker(\pi) = \operatorname{im}(\iota)$ for each $i$, i.e., $(b_i) \in \operatorname{im}(\prod \iota)$. Injectivity of $\prod \iota$ and surjectivity of $\prod \pi$ follow by the same componentwise reduction to the given exact sequence $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$. Since $\Phi$ is an isomorphism, the original $\operatorname{Hom}$ sequence is exact. [/claim] [proof] Proved inline above. [/proof] Returning to the main argument, the decomposition $F \cong P \oplus Q$ gives a natural isomorphism of functors \begin{align*} \operatorname{Hom}_R(F, -) \cong \operatorname{Hom}_R(P \oplus Q, -) \cong \operatorname{Hom}_R(P, -) \oplus \operatorname{Hom}_R(Q, -), \end{align*} where the second isomorphism sends $\varphi \in \operatorname{Hom}_R(P \oplus Q, M)$ to $(\varphi|_P, \varphi|_Q) \in \operatorname{Hom}_R(P, M) \oplus \operatorname{Hom}_R(Q, M)$, and is natural in $M$. Applying this to the short exact sequence $0 \to A \to B \to C \to 0$, the exact sequence \begin{align*} 0 \to \operatorname{Hom}_R(F, A) \to \operatorname{Hom}_R(F, B) \to \operatorname{Hom}_R(F, C) \to 0 \end{align*} decomposes as a direct sum of two sequences: \begin{align*} &0 \to \operatorname{Hom}_R(P, A) \to \operatorname{Hom}_R(P, B) \to \operatorname{Hom}_R(P, C) \to 0, \\ &0 \to \operatorname{Hom}_R(Q, A) \to \operatorname{Hom}_R(Q, B) \to \operatorname{Hom}_R(Q, C) \to 0. \end{align*} A direct sum of two sequences of abelian groups is exact if and only if each summand is exact. Since the total sequence is exact (by the claim), the first summand is exact. This is the desired exactness of $\operatorname{Hom}_R(P, -)$.[/step]

[guided]We assume $F \cong P \oplus Q$ with $F$ a [free module](/page/Free%20Module), and we must show that the functor $\operatorname{Hom}_R(P, -)$ is [exact](/page/Exact%20Sequence). The strategy has two stages: first establish that $\operatorname{Hom}_R(F, -)$ is exact for any free module $F$, then use the direct-sum decomposition $F \cong P \oplus Q$ to "project out" the $P$-component and conclude that $\operatorname{Hom}_R(P, -)$ inherits exactness. **Stage 1: Why is $\operatorname{Hom}_R(F, -)$ exact?** Let $F = \bigoplus_{i \in I} R$ be a free $R$-module with basis $\{e_i\}_{i \in I}$. The fundamental property of a free module is that an $R$-module homomorphism out of $F$ is uniquely determined by where it sends the basis elements. Formally, for any $R$-module $M$, define the evaluation map \begin{align*} \Phi_M: \operatorname{Hom}_R(F, M) &\to \prod_{i \in I} M \\ \varphi &\mapsto (\varphi(e_i))_{i \in I}. \end{align*} We claim $\Phi_M$ is an isomorphism of abelian groups. It is injective because if $\varphi(e_i) = 0$ for every $i \in I$, then $\varphi$ vanishes on the basis and hence on all of $F$ by $R$-linearity. It is surjective because given any family $(m_i)_{i \in I} \in \prod_{i \in I} M$, we can define $\varphi: F \to M$ by $\varphi((r_i)_{i \in I}) = \sum_{i \in I} r_i m_i$ (the sum is finite since elements of $F$ have finitely many nonzero components), and this $\varphi$ is the unique $R$-module homomorphism with $\varphi(e_i) = m_i$. Moreover, $\Phi$ is natural in $M$: for any [R-module homomorphism](/page/Module%20Homomorphism) $f: M \to N$, the diagram \begin{align*} \operatorname{Hom}_R(F, M) &\xrightarrow{\Phi_M} \prod_{i \in I} M \\ \downarrow f_* \quad &\qquad \downarrow \prod f \\ \operatorname{Hom}_R(F, N) &\xrightarrow{\Phi_N} \prod_{i \in I} N \end{align*} commutes, where $f_*(\varphi) = f \circ \varphi$ and $(\prod f)((m_i)) = (f(m_i))$. Commutativity holds because $(\Phi_N \circ f_*)(\varphi) = (f(\varphi(e_i)))_{i \in I} = ((\prod f) \circ \Phi_M)(\varphi)$. Now consider the [short exact sequence](/page/Exact%20Sequence) $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$. Applying the natural isomorphism $\Phi$ converts the sequence \begin{align*} 0 \to \operatorname{Hom}_R(F, A) \xrightarrow{\iota_*} \operatorname{Hom}_R(F, B) \xrightarrow{\pi_*} \operatorname{Hom}_R(F, C) \to 0 \end{align*} into the sequence \begin{align*} 0 \to \prod_{i \in I} A \xrightarrow{\prod \iota} \prod_{i \in I} B \xrightarrow{\prod \pi} \prod_{i \in I} C \to 0. \end{align*} Why is this product sequence exact? A product (i.e., direct product) of sequences of abelian groups is exact if and only if each component sequence is exact. To see this, note that the product $\prod_{i \in I} A$ is the set of all $I$-tuples, and the maps $\prod \iota$ and $\prod \pi$ act componentwise: $(\prod \iota)((a_i)) = (\iota(a_i))$ and $(\prod \pi)((b_i)) = (\pi(b_i))$. Therefore, a tuple $(b_i)$ lies in $\ker(\prod \pi)$ if and only if $\pi(b_i) = 0$ for each $i$, i.e., $b_i \in \ker(\pi) = \operatorname{im}(\iota)$ for each $i$, i.e., $(b_i) \in \operatorname{im}(\prod \iota)$. The same componentwise argument establishes injectivity of $\prod \iota$ and surjectivity of $\prod \pi$. Since $\Phi$ is a natural isomorphism, the original $\operatorname{Hom}_R(F, -)$ sequence inherits exactness. **Stage 2: Extracting exactness for $P$.** The decomposition $F \cong P \oplus Q$ induces a natural isomorphism of functors \begin{align*} \operatorname{Hom}_R(F, -) \cong \operatorname{Hom}_R(P \oplus Q, -) \cong \operatorname{Hom}_R(P, -) \oplus \operatorname{Hom}_R(Q, -). \end{align*} Concretely, an $R$-module homomorphism $\varphi: P \oplus Q \to M$ is determined by its restrictions to $P$ and $Q$, giving the isomorphism $\varphi \mapsto (\varphi|_P, \varphi|_Q)$. The inverse sends a pair $(f, g) \in \operatorname{Hom}_R(P, M) \oplus \operatorname{Hom}_R(Q, M)$ to the map $(p, q) \mapsto f(p) + g(q)$. This isomorphism is natural in $M$ because post-composition with any $R$-module homomorphism $h: M \to N$ distributes over the restriction: $(h \circ \varphi)|_P = h \circ (\varphi|_P)$. Now apply this decomposition to the short exact sequence $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$. The exact sequence $0 \to \operatorname{Hom}_R(F, A) \to \operatorname{Hom}_R(F, B) \to \operatorname{Hom}_R(F, C) \to 0$ decomposes as a direct sum of two sequences: \begin{align*} &0 \to \operatorname{Hom}_R(P, A) \to \operatorname{Hom}_R(P, B) \to \operatorname{Hom}_R(P, C) \to 0, \\ &0 \to \operatorname{Hom}_R(Q, A) \to \operatorname{Hom}_R(Q, B) \to \operatorname{Hom}_R(Q, C) \to 0. \end{align*} A [direct sum](/page/Direct%20Sum) of two sequences of abelian groups is exact if and only if each summand is exact. This follows because kernels and images in a direct sum decompose componentwise: for maps $f: A_1 \oplus A_2 \to B_1 \oplus B_2$ of the form $f = f_1 \oplus f_2$, we have $\ker(f) = \ker(f_1) \oplus \ker(f_2)$ and $\operatorname{im}(f) = \operatorname{im}(f_1) \oplus \operatorname{im}(f_2)$. Therefore $\ker(f) = \operatorname{im}(g)$ in the total sequence if and only if $\ker(f_1) = \operatorname{im}(g_1)$ and $\ker(f_2) = \operatorname{im}(g_2)$ in the respective summands. Since the total sequence is exact (by Stage 1), the $P$-summand \begin{align*} 0 \to \operatorname{Hom}_R(P, A) \to \operatorname{Hom}_R(P, B) \to \operatorname{Hom}_R(P, C) \to 0 \end{align*} is exact, as required.[/guided]

[step:Read off the lifting property from surjectivity of $\pi_*$ — $(3) \Rightarrow (1)$]Let $\pi: B \to C$ be a surjective $R$-module homomorphism and let $f: P \to C$ be any $R$-module homomorphism. We must find $\tilde{f}: P \to B$ with $\pi \circ \tilde{f} = f$. Define $A := \ker(\pi)$ and let $\iota: A \hookrightarrow B$ denote the inclusion. Then $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$ is a [short exact sequence](/page/Exact%20Sequence) of $R$-modules (exactness at $A$ holds because $\iota$ is injective, exactness at $B$ because $\ker(\pi) = \operatorname{im}(\iota)$ by definition of $A$, and exactness at $C$ because $\pi$ is surjective). By hypothesis (3), the induced sequence \begin{align*} 0 \to \operatorname{Hom}_R(P, A) \xrightarrow{\iota_*} \operatorname{Hom}_R(P, B) \xrightarrow{\pi_*} \operatorname{Hom}_R(P, C) \to 0 \end{align*} is exact. In particular, $\pi_*: \operatorname{Hom}_R(P, B) \to \operatorname{Hom}_R(P, C)$ is surjective, where $\pi_*$ is the post-composition map $\pi_*(\varphi) = \pi \circ \varphi$. Since $f \in \operatorname{Hom}_R(P, C)$, surjectivity of $\pi_*$ yields $\tilde{f} \in \operatorname{Hom}_R(P, B)$ with $\pi_*(\tilde{f}) = f$, i.e., $\pi \circ \tilde{f} = f$. This is precisely the lifting property, so $P$ is projective.[/step]

[guided]This is the most direct implication. We are given that $\operatorname{Hom}_R(P, -)$ is exact and must show that $P$ has the lifting property. Let $\pi: B \to C$ be surjective and $f: P \to C$ an $R$-module homomorphism. We need to "lift" $f$ through $\pi$: find $\tilde{f}: P \to B$ with $\pi \circ \tilde{f} = f$. Set $A := \ker(\pi)$ and let $\iota: A \hookrightarrow B$ be the inclusion. Then $0 \to A \xrightarrow{\iota} B \xrightarrow{\pi} C \to 0$ is a [short exact sequence](/page/Exact%20Sequence). (We verify: $\iota$ is injective by definition of inclusion; $\operatorname{im}(\iota) = A = \ker(\pi)$ by definition; $\pi$ is surjective by hypothesis.) Applying the exact functor $\operatorname{Hom}_R(P, -)$, we obtain the exact sequence \begin{align*} 0 \to \operatorname{Hom}_R(P, A) \xrightarrow{\iota_*} \operatorname{Hom}_R(P, B) \xrightarrow{\pi_*} \operatorname{Hom}_R(P, C) \to 0. \end{align*} What does exactness at the right end tell us? Exactness at $\operatorname{Hom}_R(P, C)$ means $\pi_*$ is surjective: every $R$-module homomorphism $f: P \to C$ equals $\pi \circ \tilde{f}$ for some $\tilde{f}: P \to B$. This is word-for-word the projective lifting property. Note where the hypothesis is consumed: $\operatorname{Hom}_R(P, -)$ is always [left-exact](/theorems/???) (for any module $P$), meaning the sequence $0 \to \operatorname{Hom}_R(P, A) \to \operatorname{Hom}_R(P, B) \to \operatorname{Hom}_R(P, C)$ is always exact. The content of condition (3) is the surjectivity of $\pi_*$ — this right-exactness is the precisely the lifting property in disguise.[/guided]