[proofplan] We verify that the formula $F(\sum [a_n] p^n) = \sum [f(a_n)] p^n$ defines a ring homomorphism $A \to B$, then prove uniqueness by showing any lift $\psi$ must satisfy $\psi([a]) = [f(a)]$ for all $a \in A/pA$. Well-definedness and continuity follow from the Witt expansion. The ring homomorphism property reduces to showing that $F$ preserves addition and multiplication, which follows from the fact that the ring operations in a strict $p$-ring are completely determined by the Teichmüller representatives and the ring structure of the residue ring. Uniqueness uses the Teichmüller construction and $p$-adic continuity. [/proofplan] [step:Define $F$ and verify well-definedness] By the [Witt Expansion in Strict $p$-Rings](/theorems/???), every element $a \in A$ has a unique representation $a = \sum_{n=0}^\infty [a_n]_A p^n$ with $a_n \in A/pA$. Define \begin{align*} F: A &\to B \\ \sum_{n=0}^\infty [a_n]_A p^n &\mapsto \sum_{n=0}^\infty [f(a_n)]_B p^n. \end{align*} Since $f(a_n) \in B/pB$ and $B$ is a strict $p$-ring, the [Witt Expansion](/theorems/???) guarantees that $\sum_{n=0}^\infty [f(a_n)]_B p^n$ converges in $B$ (each Teichmüller lift $[f(a_n)]_B$ lies in $\mathcal{O}_B = B$, and $|[f(a_n)]_B p^n| \leq |p|^n \to 0$). By uniqueness of the Witt expansion in $A$, $F$ is well-defined. [/step] [step:Verify the reduction property $F \equiv f \pmod{p}$] For any $a \in A$, write $a = [a_0]_A + p \cdot (\text{higher terms})$. Then $F(a) = [f(a_0)]_B + p \cdot (\text{higher terms})$. Reducing modulo $p$: the image of $F(a)$ in $B/pB$ is $f(a_0)$, and the image of $a$ in $A/pA$ is $a_0$. So the reduction of $F$ modulo $p$ is $a_0 \mapsto f(a_0) = f(a_0)$, which is exactly $f$. Thus $F \equiv f \pmod{p}$. [/step] [step:Show that $F$ is a ring homomorphism] We verify that $F$ preserves addition and multiplication. The key observation is that $F$ is $p$-adically continuous: if $a \equiv b \pmod{p^m}$, then the first $m$ Teichmüller coefficients of $a$ and $b$ agree (by uniqueness of the Witt expansion), so $F(a) \equiv F(b) \pmod{p^m}$. **Multiplicativity.** For $a, b \in A$ with Witt expansions $a = \sum [a_n]_A p^n$ and $b = \sum [b_n]_A p^n$, consider the product $ab$. Modulo $p$, $ab \equiv [a_0]_A [b_0]_A = [a_0 b_0]_A \pmod{p}$ (using multiplicativity of the [Teichmüller Map](/theorems/???)). So the zeroth Witt coefficient of $ab$ is $a_0 b_0$. Similarly, $F(a) F(b) \equiv [f(a_0)]_B [f(b_0)]_B = [f(a_0) f(b_0)]_B = [f(a_0 b_0)]_B \pmod{p}$. Since $f$ is a ring homomorphism, $f(a_0 b_0) = f(a_0) f(b_0)$, so $F(ab) \equiv F(a)F(b) \pmod{p}$. For the general congruence modulo $p^m$: the Witt coefficients of $ab$ are universal polynomials (the Witt multiplication polynomials) in the coefficients $a_0, \ldots, a_{m-1}, b_0, \ldots, b_{m-1}$ with coefficients in $\mathbb{Z}$. Since $f$ is a ring homomorphism on $A/pA$ and these polynomials are defined over $\mathbb{Z}$, applying $f$ to each coefficient commutes with the polynomial operations. Hence $F(ab) \equiv F(a)F(b) \pmod{p^m}$ for all $m$. Since $B$ is $p$-adically separated ($\bigcap_{m} p^m B = \{0\}$, which follows from $p$-torsion-freeness and completeness), $F(ab) = F(a)F(b)$. **Additivity.** The same argument applies with the Witt addition polynomials replacing the multiplication polynomials. The zeroth coefficient of $a + b$ satisfies the reduction $(a + b) \bmod p = a_0 + b_0$, and $f(a_0 + b_0) = f(a_0) + f(b_0)$, so $F(a + b) \equiv F(a) + F(b) \pmod{p}$. The general congruence modulo $p^m$ follows from the same universality argument. [guided] The ring homomorphism property of $F$ is subtle because the Witt addition and multiplication are not simply termwise operations on the Teichmüller coefficients. The $n$-th Witt coefficient of $a + b$ depends on $a_0, \ldots, a_n$ and $b_0, \ldots, b_n$ via a universal polynomial $S_n$ over $\mathbb{Z}$. Similarly, the $n$-th coefficient of $ab$ is given by a universal polynomial $P_n$. The crucial insight is that these polynomials $S_n$ and $P_n$ are the same regardless of which strict $p$-ring we work in -- they are "universal" in the sense that they depend only on the prime $p$. Applying $f$ to the inputs of these polynomials commutes with evaluating the polynomials (since $f$ is a ring homomorphism and the polynomials have $\mathbb{Z}$-coefficients). Concretely: if the $n$-th Witt coefficient of $a + b$ in $A$ is $S_n(a_0, \ldots, a_n, b_0, \ldots, b_n)$, then the $n$-th coefficient of $F(a) + F(b)$ in $B$ is $S_n(f(a_0), \ldots, f(a_n), f(b_0), \ldots, f(b_n)) = f(S_n(a_0, \ldots, a_n, b_0, \ldots, b_n))$ (the last equality uses that $f$ is a ring homomorphism). This is exactly the $n$-th coefficient of $F(a + b)$. [/guided] [/step] [step:Prove uniqueness of the lift] Let $\psi: A \to B$ be any ring homomorphism with $\psi \equiv f \pmod{p}$. We show $\psi = F$. Since $\psi$ is a ring homomorphism and $\psi(p \cdot 1_A) = p \cdot 1_B$ (as $\psi(1) = 1$ and $\psi$ preserves addition), $\psi$ is $p$-adically continuous: $\psi(p^m A) \subseteq p^m B$. It suffices to show $\psi([a]_A) = [f(a)]_B$ for all $a \in A/pA$, because then $\psi(\sum [a_n]_A p^n) = \sum \psi([a_n]_A) p^n = \sum [f(a_n)]_B p^n = F(\sum [a_n]_A p^n)$. Fix $a \in A/pA$. Since $A/pA$ is perfect, for each $n \geq 0$ there exists $\alpha_n \in A$ lifting $a^{p^{-n}} \in A/pA$. By the [Teichmüller Map](/theorems/???) construction, $[a]_A = \lim_{n \to \infty} \alpha_n^{p^n}$. Since $\psi$ is a ring homomorphism, $\psi(\alpha_n^{p^n}) = \psi(\alpha_n)^{p^n}$. The element $\psi(\alpha_n) \in B$ lifts $f(a^{p^{-n}}) = f(a)^{p^{-n}} \in B/pB$ (since $\psi \equiv f \pmod{p}$ and $f$ is a ring homomorphism). By the Teichmüller construction in $B$, $[f(a)]_B = \lim_{n \to \infty} \psi(\alpha_n)^{p^n}$. Therefore \begin{align*} \psi([a]_A) = \psi\!\left(\lim_{n \to \infty} \alpha_n^{p^n}\right) = \lim_{n \to \infty} \psi(\alpha_n^{p^n}) = \lim_{n \to \infty} \psi(\alpha_n)^{p^n} = [f(a)]_B. \end{align*} The second equality uses $p$-adic continuity of $\psi$. [guided] The uniqueness proof has two key ingredients: 1. **$p$-adic continuity of $\psi$**: Since $\psi$ is a ring homomorphism with $\psi(p) = p$, it preserves the $p$-adic filtration: $\psi(p^m A) \subseteq p^m B$. This means $\psi$ commutes with $p$-adic limits. 2. **Characterization of Teichmüller lifts via $p$-power limits**: The Teichmüller lift $[a]$ is the limit $\lim \alpha_n^{p^n}$ for any sequence of lifts $\alpha_n$ of $a^{p^{-n}}$. A ring homomorphism $\psi$ sends $\alpha_n^{p^n}$ to $\psi(\alpha_n)^{p^n}$, and $\psi(\alpha_n)$ is a lift of $f(a^{p^{-n}}) = f(a)^{p^{-n}}$ (here we use both $\psi \equiv f \pmod{p}$ and that $f$ is a ring homomorphism). Combining these: $\psi([a]) = \lim \psi(\alpha_n)^{p^n} = [f(a)]$. Since $\psi$ is determined on Teichmüller lifts and every element of $A$ is a $p$-adic combination of Teichmüller lifts (the Witt expansion), $\psi$ is completely determined. The formula forces $\psi = F$. [/guided] [/step]