[proofplan] We use the direct-summand characterization of finitely generated projective modules. Since $P$ is finitely generated projective, it is a direct summand of a finite free $R$-module; localizing the resulting decomposition shows that $S^{-1}P$ is a direct summand of a finite free $S^{-1}R$-module. Finally, localization preserves isomorphisms and finite direct sums, so the assignment on finitely generated projective modules is a monoid homomorphism and hence induces the asserted homomorphism on $K_0$. [/proofplan]

[step:Regard localization as a functor to $S^{-1}R$-modules] Set $A := S^{-1}R$. For every $R$-module $M$, let $S^{-1}M$ denote the localization of $M$, regarded as an $A$-module by \begin{align*} \frac{r}{s}\cdot \frac{m}{t} := \frac{rm}{st} \end{align*} for $r \in R$, $m \in M$, and $s,t \in S$. If $f:M\to N$ is an $R$-[linear map](/page/Linear%20Map), define \begin{align*} S^{-1}f:S^{-1}M&\to S^{-1}N \end{align*} by \begin{align*} (S^{-1}f)\left(\frac{m}{s}\right)=\frac{f(m)}{s}. \end{align*} The defining [equivalence relation](/page/Equivalence%20Relation) for localization shows that $S^{-1}f$ is well-defined and $A$-linear. Thus localization carries isomorphic $R$-modules to isomorphic $A$-modules. [/step]

[step:Localize finite direct sums explicitly]Let $M$ and $N$ be $R$-modules. Define an $A$-linear map \begin{align*} \Theta_{M,N}:S^{-1}(M\oplus N)&\to S^{-1}M\oplus S^{-1}N \end{align*} by \begin{align*} \Theta_{M,N}\left(\frac{(m,n)}{s}\right)=\left(\frac{m}{s},\frac{n}{s}\right). \end{align*} Define \begin{align*} \Psi_{M,N}:S^{-1}M\oplus S^{-1}N&\to S^{-1}(M\oplus N) \end{align*} by \begin{align*} \Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)=\frac{(tm,sn)}{st}. \end{align*} These maps are well-defined because replacing a fraction in a localization only multiplies the numerator by an element of $S$, and the displayed formula uses a common denominator. A direct substitution gives \begin{align*} \Theta_{M,N}\circ \Psi_{M,N}=\operatorname{id}_{S^{-1}M\oplus S^{-1}N} \end{align*} and \begin{align*} \Psi_{M,N}\circ \Theta_{M,N}=\operatorname{id}_{S^{-1}(M\oplus N)}. \end{align*} Therefore \begin{align*} S^{-1}(M\oplus N)\cong S^{-1}M\oplus S^{-1}N \end{align*} as $A$-modules.[/step]

[guided]The point of this step is to make the compatibility with direct sums completely explicit, because this compatibility is what later makes localization descend to $K_0$. Let $M$ and $N$ be $R$-modules. We define \begin{align*} \Theta_{M,N}:S^{-1}(M\oplus N)\to S^{-1}M\oplus S^{-1}N \end{align*} by \begin{align*} \Theta_{M,N}\left(\frac{(m,n)}{s}\right)=\left(\frac{m}{s},\frac{n}{s}\right). \end{align*} This map separates the two components after localization. Conversely, if two localized elements have different denominators, we combine them by passing to a common denominator. Define \begin{align*} \Psi_{M,N}:S^{-1}M\oplus S^{-1}N\to S^{-1}(M\oplus N) \end{align*} by \begin{align*} \Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)=\frac{(tm,sn)}{st}. \end{align*} The denominator $st$ lies in $S$ because $S$ is multiplicative. The numerator $(tm,sn)$ lies in $M\oplus N$ because $M$ and $N$ are $R$-modules. We now verify that these two maps are inverse. Starting from $\left(\frac{m}{s},\frac{n}{t}\right)$, we compute \begin{align*} \Theta_{M,N}\left(\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)\right)=\Theta_{M,N}\left(\frac{(tm,sn)}{st}\right). \end{align*} Hence \begin{align*} \Theta_{M,N}\left(\Psi_{M,N}\left(\frac{m}{s},\frac{n}{t}\right)\right)=\left(\frac{tm}{st},\frac{sn}{st}\right)=\left(\frac{m}{s},\frac{n}{t}\right). \end{align*} Starting from $\frac{(m,n)}{s}$, we compute \begin{align*} \Psi_{M,N}\left(\Theta_{M,N}\left(\frac{(m,n)}{s}\right)\right)=\Psi_{M,N}\left(\frac{m}{s},\frac{n}{s}\right). \end{align*} Therefore \begin{align*} \Psi_{M,N}\left(\Theta_{M,N}\left(\frac{(m,n)}{s}\right)\right)=\frac{(sm,sn)}{s^2}=\frac{(m,n)}{s}. \end{align*} Thus $S^{-1}(M\oplus N)\cong S^{-1}M\oplus S^{-1}N$ as $A$-modules.[/guided]

[step:Identify the localization of a finite free module] For every integer $n\ge 0$, repeated application of the preceding direct-sum isomorphism gives \begin{align*} S^{-1}(R^n)\cong (S^{-1}R)^n=A^n \end{align*} as $A$-modules. In particular, localization sends finite free $R$-modules to finite free $A$-modules of the same rank. [/step]

[step:Localize a finite free summand decomposition of $P$] Since $P$ is finitely generated projective, there exist an $R$-module $Q$, an integer $n\ge 0$, and an $R$-module isomorphism \begin{align*} \alpha:P\oplus Q\to R^n. \end{align*} Localizing $\alpha$ gives an $A$-module isomorphism \begin{align*} S^{-1}\alpha:S^{-1}(P\oplus Q)\to S^{-1}(R^n). \end{align*} Using the direct-sum and finite-free identifications from the previous steps, we obtain an $A$-module isomorphism \begin{align*} S^{-1}P\oplus S^{-1}Q\cong A^n. \end{align*} Thus $S^{-1}P$ is a direct summand of the finite free $A$-module $A^n$. By the direct-summand characterization of finitely generated projective modules, $S^{-1}P$ is finitely generated projective as an $A=S^{-1}R$-module. [/step]

[step:Show that localization descends to $K_0$] Let $V(R)$ be the commutative monoid of isomorphism classes of finitely generated projective $R$-modules under direct sum, and let $V(A)$ be the corresponding monoid for $A=S^{-1}R$. Define \begin{align*} \ell:V(R)\to V(A) \end{align*} by \begin{align*} \ell([P])=[S^{-1}P]. \end{align*} This is well-defined because localization sends isomorphic $R$-modules to isomorphic $A$-modules, and the preceding step shows that $S^{-1}P$ is finitely generated projective whenever $P$ is. For finitely generated projective $R$-modules $P$ and $P'$, the direct-sum isomorphism gives \begin{align*} \ell([P]+[P'])=\ell([P\oplus P'])=[S^{-1}(P\oplus P')]. \end{align*} Hence \begin{align*} \ell([P]+[P'])=[S^{-1}P\oplus S^{-1}P']=\ell([P])+\ell([P']). \end{align*} Also $\ell([0])=[0]$. Therefore $\ell$ is a monoid homomorphism. Since $K_0(R)$ and $K_0(A)$ are the group completions of $V(R)$ and $V(A)$, respectively, $\ell$ induces a unique [group homomorphism](/page/Group%20Homomorphism) \begin{align*} K_0(R)\to K_0(A) \end{align*} sending $[P]$ to $[S^{-1}P]$. Replacing $A$ by $S^{-1}R$ gives the asserted homomorphism \begin{align*} K_0(R)\to K_0(S^{-1}R), \qquad [P]\mapsto [S^{-1}P]. \end{align*} This completes the proof. [/step]