Fix distinct $x_1, x_2, ..., x_n, x_{n+1}$. Let $T:P_n(\mathbb R) \rightarrow \mathbb R^{n+1}$ be defined by $T(f)=(f(x_1), f(x_2). ..., f(x_n), f(x_{n+1}))$. Prove that T is an isomorphism.
So basically, I'm trying to prove that there is an morphism $G:\mathbb R^{n+1} \rightarrow P_n(\mathbb R)$. My first idea is that because we have n+1 values of this polynomial, it can be represented in a unique expression $f=(c_1x_1-r_1)*(c_2x_2-r_2)...(c_nx_n-r_n)(c_{n+1}x_{n+1}-r_{n+1})$ where $c_i,r_i \in \mathbb R$ for all $i \in \{1,2,...,n+1\}$
How do I justify my first claim (Assuming that is correct)? And, once I have this unique polynomial, how do I go back to $P_n(\mathbb R)$ from the polynomial?