[proofplan]
We factor the problem through the polynomial ring $R[\{T_u\}_{u \in U}]$. By the universal property of the polynomial ring, an $R$-algebra homomorphism out of $R[\{T_u\}_{u \in U}]$ is determined by the images of the indeterminates $\{T_u\}$. Such a map descends to the quotient $R_U = R[\{T_u\}_{u \in U}]/I_U$ if and only if it kills the ideal $I_U$, which forces each $T_u$ to map to $f(u)^{-1}$. Since $f(u)$ is a unit in $A$ by hypothesis, the images are uniquely determined, giving both existence and uniqueness of $h$.
[/proofplan]
[step:Lift the problem to the polynomial ring $R[\{T_u\}_{u \in U}]$ via its universal property]
The ring $R_U$ is defined as the quotient $R[\{T_u\}_{u \in U}]/I_U$, where $I_U$ is the ideal generated by $\{uT_u - 1 : u \in U\}$. Let $\pi: R[\{T_u\}_{u \in U}] \to R_U$ denote the canonical projection, so that $\iota = \pi \circ j$ where $j: R \hookrightarrow R[\{T_u\}_{u \in U}]$ is the natural inclusion.
By the [Universal Property of Polynomial Algebra](/theorems/2820), for any $R$-algebra $A$ and any choice of elements $\{a_u\}_{u \in U}$ in $A$, there is a unique $R$-algebra homomorphism
\begin{align*}
\Phi: R[\{T_u\}_{u \in U}] &\to A \\
T_u &\mapsto a_u.
\end{align*}
Being an $R$-algebra homomorphism means $\Phi(r) = f(r)$ for all $r \in R$, where $A$ is regarded as an $R$-algebra via $f$.
[guided]
The polynomial ring $R[\{T_u\}_{u \in U}]$ is the free commutative $R$-algebra on the set $\{T_u : u \in U\}$. Its universal property states that giving an $R$-algebra homomorphism out of it is the same as choosing where each indeterminate goes — there are no relations to respect. This makes the polynomial ring the natural first step: we build a map on the free object and then check whether it descends to the quotient.
The key point is that $\Phi$ automatically satisfies $\Phi \circ j = f$ (where $j: R \hookrightarrow R[\{T_u\}_{u \in U}]$ is the inclusion), because $\Phi$ is an $R$-algebra homomorphism and $A$ is an $R$-algebra via $f$. In particular, for $r \in R$, we have $\Phi(j(r)) = \Phi(r) = f(r)$.
[/guided]
[/step]
[step:Determine the images $a_u = f(u)^{-1}$ by requiring $\Phi(I_U) = 0$]
The map $\Phi$ descends to the quotient $R_U = R[\{T_u\}_{u \in U}]/I_U$ if and only if $I_U \subseteq \ker \Phi$. Since $I_U$ is generated by the elements $\{uT_u - 1 : u \in U\}$, the condition $I_U \subseteq \ker \Phi$ is equivalent to
\begin{align*}
\Phi(uT_u - 1) = 0 \quad \text{for every } u \in U.
\end{align*}
Computing, $\Phi(uT_u - 1) = f(u) \cdot a_u - 1$, where $a_u = \Phi(T_u)$ and we used that $\Phi$ is an $R$-algebra homomorphism so $\Phi(u) = f(u)$. The equation $f(u) \cdot a_u - 1 = 0$ has a solution in $A$ if and only if $f(u)$ is a unit in $A$, in which case the unique solution is
\begin{align*}
a_u = f(u)^{-1}.
\end{align*}
Since $f(u)$ is a unit in $A$ for every $u \in U$ by hypothesis, the choice $a_u = f(u)^{-1}$ is forced and is the only possibility.
[guided]
This is where the hypothesis that $f(u)$ is a unit in $A$ is consumed. Without it, the equation $f(u) \cdot a_u = 1$ may have no solution in $A$, and the map would not descend to $R_U$. With it, the equation has a unique solution $a_u = f(u)^{-1}$, which establishes both existence and uniqueness simultaneously.
Note that uniqueness of $a_u$ is automatic from the invertibility of $f(u)$: if $f(u) \cdot a_u = 1 = f(u) \cdot a_u'$, then multiplying by $f(u)^{-1}$ gives $a_u = a_u'$.
[/guided]
[/step]
[step:Conclude existence and uniqueness of $h: R_U \to A$]
Define $\Phi: R[\{T_u\}_{u \in U}] \to A$ to be the unique $R$-algebra homomorphism sending $T_u \mapsto f(u)^{-1}$ for each $u \in U$. By the computation in the previous step, $\Phi(uT_u - 1) = f(u) \cdot f(u)^{-1} - 1 = 0$ for every $u \in U$, so $I_U \subseteq \ker \Phi$.
By the universal property of quotient rings, $\Phi$ factors uniquely through $\pi$: there is a unique ring homomorphism $h: R_U \to A$ satisfying $h \circ \pi = \Phi$. Explicitly,
\begin{align*}
h(p + I_U) = \Phi(p)
\end{align*}
for every $p \in R[\{T_u\}_{u \in U}]$, where the right-hand side is the evaluation of the polynomial $p$ obtained by substituting $f(u)^{-1}$ for each $T_u$ and applying $f$ to the coefficients. We verify the compatibility condition: for every $r \in R$,
\begin{align*}
(h \circ \iota)(r) = h(\pi(j(r))) = \Phi(j(r)) = f(r),
\end{align*}
so $h \circ \iota = f$.
**Uniqueness of $h$:** Suppose $h': R_U \to A$ is any ring homomorphism with $h' \circ \iota = f$. Define $\Phi' := h' \circ \pi: R[\{T_u\}_{u \in U}] \to A$. Then $\Phi'$ is an $R$-algebra homomorphism (since $\Phi' \circ j = h' \circ \pi \circ j = h' \circ \iota = f$) and $I_U \subseteq \ker \Phi'$ (since $\Phi'$ factors through $\pi$). By the previous step, $\Phi'$ must send $T_u \mapsto f(u)^{-1}$ for each $u$. Since the universal property of the polynomial ring determines an $R$-algebra homomorphism uniquely from the images of the indeterminates, $\Phi' = \Phi$. Therefore $h' \circ \pi = h \circ \pi$, and since $\pi$ is surjective, $h' = h$.
[guided]
The argument has two components. **Existence**: the universal property of the polynomial ring gives us a map $\Phi$, the vanishing on generators of $I_U$ shows $I_U \subseteq \ker \Phi$, and the universal property of quotient rings gives the induced map $h$. The compatibility $h \circ \iota = f$ follows by tracing the composition through the factorisation $\iota = \pi \circ j$.
**Uniqueness**: any other $h'$ with $h' \circ \iota = f$ lifts to a map $\Phi' = h' \circ \pi$ on the polynomial ring. Since $\Phi'$ is an $R$-algebra homomorphism that kills $I_U$, the analysis of the previous step forces $\Phi'(T_u) = f(u)^{-1}$ for every $u$. The universal property of the polynomial ring then forces $\Phi' = \Phi$, so $h' = h$.
The surjectivity of $\pi$ is essential in the final step: from $h' \circ \pi = h \circ \pi$ we can only conclude $h' = h$ because every element of $R_U$ is in the image of $\pi$.
[/guided]
[/step]
