[proofplan] The quotient map $p:P\to P''$ splits because $P''$ is projective. Exactness then identifies $P'$ with the kernel of $p$, and the splitting gives an isomorphism $P\cong P'\oplus P''$. Finally, the operation defining $V(R)$ and hence $K_0(R)$ is induced by direct sum, so this isomorphism gives the desired equality of classes. [/proofplan] [step:Split the surjection onto the projective quotient] Since the sequence is exact, the map \begin{align*}p:P\longrightarrow P''\end{align*} is a surjective homomorphism of left $R$-modules. Since $P''$ is projective, the surjection $p$ admits an $R$-linear section. Equivalently, there is an $R$-[linear map](/page/Linear%20Map) \begin{align*}s:P''\longrightarrow P\end{align*} such that \begin{align*}p\circ s=\operatorname{id}_{P''}.\end{align*} This is the splitting assertion for a surjection onto a projective module, as in [citetheorem:8640]. [guided] The only structural input is the projectivity of the quotient module $P''$. Exactness of \begin{align*}0 \longrightarrow P' \xrightarrow{i} P \xrightarrow{p} P'' \longrightarrow 0\end{align*} says, in particular, that $p:P\to P''$ is surjective. The defining lifting property of a projective module says that whenever a projective module maps to the target of a surjection, that map lifts through the surjection. Apply this property to the identity map \begin{align*}\operatorname{id}_{P''}:P''\longrightarrow P''\end{align*} and to the surjection $p:P\to P''$. We obtain an $R$-linear map \begin{align*}s:P''\longrightarrow P\end{align*} with \begin{align*}p\circ s=\operatorname{id}_{P''}.\end{align*} Thus $s$ is a section of $p$. This is precisely the splitting step recorded in [citetheorem:8640]: a short exact sequence whose quotient is projective splits. [/guided] [/step] [step:Use the splitting to identify the middle module with a direct sum] Define an $R$-linear map \begin{align*}\Phi:P'\oplus P''\longrightarrow P\end{align*} by \begin{align*}\Phi(x,y)=i(x)+s(y)\end{align*} for $x\in P'$ and $y\in P''$. We prove that $\Phi$ is an isomorphism. First suppose $\Phi(x,y)=0$. Applying $p$ gives \begin{align*}0=p(\Phi(x,y))=p(i(x))+p(s(y)).\end{align*} Exactness gives $\operatorname{im} i=\ker p$, so $p\circ i=0$. Since $p\circ s=\operatorname{id}_{P''}$, the preceding equality becomes \begin{align*}0=y.\end{align*} Then $i(x)=0$, and exactness at $P'$ gives that $i$ is injective, so $x=0$. Hence $\Phi$ is injective. Now let $z\in P$. Define \begin{align*}y:=p(z)\in P''\end{align*} and \begin{align*}w:=z-s(y)\in P.\end{align*} Then \begin{align*}p(w)=p(z)-p(s(y))=y-y=0,\end{align*} so $w\in\ker p$. Since $\ker p=\operatorname{im} i$, there exists $x\in P'$ such that $i(x)=w$. Therefore \begin{align*}z=i(x)+s(y)=\Phi(x,y),\end{align*} so $\Phi$ is surjective. Thus \begin{align*}P\cong P'\oplus P''\end{align*} as left $R$-modules. [/step] [step:Pass from the direct sum decomposition to the equality in $K_0(R)$] By [citetheorem:8629], $V(R)$ is the commutative monoid of isomorphism classes of finitely generated projective left $R$-modules under direct sum, so the isomorphism $P\cong P'\oplus P''$ gives \begin{align*} [P]=[P'\oplus P'']=[P']+[P''] \end{align*} in $V(R)$. By the group-completion construction recalled in [citetheorem:8633], the canonical map from $V(R)$ to its group completion $K_0(R)$ is a monoid homomorphism, so it preserves this equality. Hence, in $K_0(R)$, \begin{align*} [P]=[P']+[P'']. \end{align*} This proves the theorem. [/step]