Showing that $R[x_1,...,x_n]$ is the only $R$-algebra up to isomorphism that satisfies the universal property of the polynomial ring. I am currently working through some exercises in Computational Commutative Algebra 1, and I have run in to some difficulties.
The universal property of the polynomial ring states:
Let $S$ be an $R$-algebra ($R$ a commutative ring) with structural homomorphism $\phi:R \to S$. For $n \in \mathbb{N}$ and elements $s_1,\dots,s_n$, there exists a unique homomorphism: $\psi: R[x_1,\ldots,x_n] \to S$ with $\psi(x_i)=s_i$ and with $\psi(r)=\phi(r)$ for all $r \in R$.
I am trying to prove that any $R$-algebra satisfying this property is isomorphic to $R[x_1,\ldots,x_n]$ with a unique isomorphism, but i am running in to some problems since l'm new to $R$-algebras.
What I have tried so far:
Assume that $T$ is an $R$-algebra with structural homomorphism $\phi_T: R \to T$. Let $t_1,ldots,t_n \in T$ and assume that there exists a unique homomorphism $\psi_T : T \to S$ such that $\psi_T(t_i)=s_i$. I want to construct a unique $R$-algebra homomorphism $\Phi: R[x_1,\ldots,x_n] \to T$. My first thought was the homomorphism induced by $x_i \mapsto t_i$. Is this enough information to define a unique homomorphism? 
I also have some difficulty figuring out how the universal property comes into play. I think that it is used in proving the uniqueness of the $R$-algebra isomorphism between $R[x_1,\ldots,x_n]$ and $T$, but I am not sure.
Any help would be appreciated.
 A: So the universal property is as follows:
For every $R$-Algebra $\Phi : R \to S$ and for elements $s_1, \ldots, s_n \in S$ there exists a unique morphism $\psi : R[x_1, \ldots x_n] \to S$ with $\psi(x_i) = s_i$.
Now let's assume another $R$-Algebra $T$ together with elements $t_1, \ldots t_n \in T$ satisfies this universal property. Then we get a unique morphism $\psi : R[x_1, \ldots, x_n] \to T$ with $\psi(x_i) = t_i$ by the universal property of the polynomial ring and we get a unique morphism $\tilde{\psi} : T \to R[x_1, \ldots, x_n]$ with $\tilde{\psi}(t_i) = x_i$ by the universal property of $T$. Now, $\psi \circ \tilde{\psi} : T \to T$ is a morphism satisfying $\psi \circ \tilde{\psi}(t_i) = t_i$ and by uniqueness of such a morphism (we use the universal propery for $T$ with $S = T$) we obtain $\psi \circ \tilde{\psi} = \text{id}_T$. Analogously one sees that $\tilde{\psi} \circ \psi : R[x_1, \ldots, x_n] \to R[x_1, \ldots, x_n]$ is equal to $\text{id}_{R[x_1, \ldots, x_n]}$. Thus $R[x_1, \ldots, x_n] \simeq T$ with unique isomorphism given by $\psi$.
In general: Objects defined by universal properties are unique up to unique isomorphism and the proof is the same as above.
