[proofplan] We use the lifting property in the definition of a projective module to lift the identity map on $P''$ through the surjection $p:P\to P''$. This produces a section $s:P''\to P$ of $p$. The existence of such a section lets us decompose every element of $P$ uniquely as a sum of an element of $\ker p$ and an element of $s(P'')$. Finally, exactness identifies $\ker p$ with $P'$ via the injective map $i$, giving the desired isomorphism $P\cong P'\oplus P''$. [/proofplan]

[step:Lift the identity map through the surjection $p$]Since the sequence is exact, the homomorphism \begin{align*} p:P\to P'' \end{align*} is surjective. Since $P''$ is projective, the defining lifting property of projective modules applied to the surjection $p:P\to P''$ and to the identity homomorphism \begin{align*} \operatorname{id}_{P''}:P''\to P'' \end{align*} gives an $R$-linear homomorphism \begin{align*} s:P''\to P \end{align*} such that \begin{align*} p\circ s=\operatorname{id}_{P''}. \end{align*} Thus $s$ is a section of $p$, and the short exact sequence splits at the level of the quotient map.[/step]

[guided]The only hypothesis not coming from exactness is the projectivity of $P''$, so we use it immediately. The definition of a projective module says that whenever an $R$-[linear map](/page/Linear%20Map) $q:M\to N$ is surjective and an $R$-linear map $f:P''\to N$ is given, there is an $R$-linear map $\widetilde f:P''\to M$ with \begin{align*} q\circ \widetilde f=f. \end{align*} In the present situation, exactness of \begin{align*} 0 \longrightarrow P' \xrightarrow{i} P \xrightarrow{p} P'' \longrightarrow 0 \end{align*} implies that $p:P\to P''$ is surjective. We take $q=p$, $M=P$, $N=P''$, and \begin{align*} f=\operatorname{id}_{P''}:P''\to P''. \end{align*} Projectivity of $P''$ therefore gives an $R$-linear homomorphism \begin{align*} s:P''\to P \end{align*} such that \begin{align*} p\circ s=\operatorname{id}_{P''}. \end{align*} This equation is exactly the statement that $s$ is a right inverse, or section, of $p$. It means that every element $y\in P''$ can be lifted to the specific element $s(y)\in P$, and applying $p$ to that lift recovers $y$.[/guided]

[step:Decompose $P$ as $\ker p$ plus the image of the section] Let \begin{align*} s(P'')=\{s(y):y\in P''\}\subset P \end{align*} denote the image submodule of $s$. We first show that \begin{align*} P=\ker p+s(P''). \end{align*} For any $x\in P$, define \begin{align*} y=p(x)\in P''. \end{align*} Then \begin{align*} x=\bigl(x-s(p(x))\bigr)+s(p(x)). \end{align*} The second summand belongs to $s(P'')$ by definition. For the first summand, $R$-linearity of $p$ and the identity $p\circ s=\operatorname{id}_{P''}$ give \begin{align*} p\bigl(x-s(p(x))\bigr)=p(x)-p(s(p(x)))=p(x)-p(x)=0. \end{align*} Hence $x-s(p(x))\in\ker p$, proving $P=\ker p+s(P'')$. Next we show that this sum is direct. If $z\in \ker p\cap s(P'')$, then there exists $y\in P''$ such that $z=s(y)$, and also $p(z)=0$. Therefore \begin{align*} 0=p(z)=p(s(y))=y. \end{align*} Thus $y=0$, so $z=s(0)=0$. Hence \begin{align*} \ker p\cap s(P'')=\{0\}. \end{align*} Therefore \begin{align*} P=\ker p\oplus s(P''). \end{align*} [/step]

[step:Identify the two direct summands with $P'$ and $P''$] Let \begin{align*} \operatorname{im} i=\{i(x'):x'\in P'\}\subset P \end{align*} denote the image submodule of $i$. Exactness at $P$ gives \begin{align*} \operatorname{im} i=\ker p. \end{align*} Exactness at $P'$ gives that \begin{align*} i:P'\to P \end{align*} is injective. Therefore the corestriction \begin{align*} i_{\ker}:P'\to \ker p \end{align*} defined by \begin{align*} i_{\ker}(x')=i(x') \end{align*} for $x'\in P'$ is an $R$-linear isomorphism. Also, the section \begin{align*} s:P''\to s(P'') \end{align*} is an $R$-linear isomorphism onto its image. Indeed, it is surjective onto $s(P'')$ by definition, and if $s(y)=0$, then applying $p$ gives \begin{align*} y=p(s(y))=p(0)=0. \end{align*} Thus $s$ is injective. Combining these identifications with \begin{align*} P=\ker p\oplus s(P'') \end{align*} gives an $R$-linear isomorphism \begin{align*} P'\oplus P''\to P \end{align*} defined by \begin{align*} (x',y)\mapsto i(x')+s(y). \end{align*} Hence \begin{align*} P\cong P'\oplus P'' \end{align*} as left $R$-modules. [/step]

[step:Apply the splitting to the finitely generated projective case] If $P'$, $P$, and $P''$ are finitely generated projective left $R$-modules, then in particular $P''$ is projective. The preceding steps apply without any additional hypothesis on $P'$ or $P$, and therefore give \begin{align*} P\cong P'\oplus P''. \end{align*} This proves the stated special case and completes the proof. [/step]