**Construction.** For $a \in R/xR$, fix for each $n \geq 0$ an arbitrary lift $\alpha_n \in R$ of $a^{p^{-n}} \in R/xR$ (which exists uniquely since $R/xR$ is perfect), so $\alpha_n \equiv a^{p^{-n}} \pmod{x}$. Define $\beta_n = \alpha_n^{p^n}$, and set $[a] = \lim_{n \to \infty} \beta_n$. **The limit exists and is independent of choices.** We show $\beta_{n+1} - \beta_n \in x^{n+1} R$. Indeed: \begin{align*} \beta_{n+1} - \beta_n = (\alpha_{n+1}^p)^{p^n} - \alpha_n^{p^n}. \end{align*} Now $\alpha_{n+1}^p \equiv (a^{p^{-n-1}})^p = a^{p^{-n}} \equiv \alpha_n \pmod{x}$. Applying the Frobenius lemma $n$ times (starting from congruence mod $x$) gives $(\alpha_{n+1}^p)^{p^n} \equiv \alpha_n^{p^n} \pmod{x^{n+1}}$. So $\beta_{n+1} - \beta_n \in x^{n+1}R$, and the $\beta_n$ form a Cauchy sequence, converging by $x$-adic completeness. Independence of choices follows: two such sequences are interlaceable to form a third respectable sequence, which must converge; so both sequences share the same limit. **The reduction.** Since $R \to R/xR$ is continuous (with discrete topology on $R/xR$), $[a] = \lim \alpha_n^{p^n} \equiv \lim (a^{p^{-n}})^{p^n} = a \pmod{x}$. **Multiplicativity.** If $b \in R/xR$ with lifts $\gamma_n \equiv b^{p^{-n}} \pmod{x}$, then $\alpha_n \gamma_n$ lifts $(ab)^{p^{-n}}$, so $[ab] = \lim (\alpha_n \gamma_n)^{p^n} = \lim \alpha_n^{p^n} \lim \gamma_n^{p^n} = [a][b]$. **Additivity when $\operatorname{char}(R) = p$.** In this case $(\alpha + \gamma)^p = \alpha^p + \gamma^p$ (all cross terms vanish), so $\alpha_n + \gamma_n$ lifts $(a+b)^{p^{-n}}$, giving $[a+b] = \lim (\alpha_n + \gamma_n)^{p^n} = [a] + [b]$. **Uniqueness.** If $\phi: R/xR \to R$ is another multiplicative section with $\phi(a) \equiv a \pmod{x}$, then $\phi(a^{p^{-n}}) \equiv a^{p^{-n}} \pmod{x}$ is a valid choice of $\alpha_n$. So $[a] = \lim \phi(a^{p^{-n}})^{p^n} = \lim \phi(a) = \phi(a)$ (the map $\phi$ is $p$-adically continuous since $\phi$ is multiplicative and thus $\phi(a^{p^n}) = \phi(a)^{p^n}$, making $\phi$ a $p$-power limit).