Let $f : \mathbb{A}^n \to \mathbb{A}$ be any function (not necessarily regular) such that the set $\{(p,f(p))~|~p \in \mathbb{A}^n \}$ is an algebraic variety in $\mathbb{A}^{n+1}$. Then projection onto the first $n$ coordinates induces a bijective morphism of algebraic varieties. Need this always be an isomorphism?
Clearly, this is equivalent to showing that $f$ must, in fact, always be regular whenever its graph $G$ is algebraic. Here's what I've been able to deduce when the base field $k$ is algebraically closed and of characteristic $0$: Since $G$ surjects onto a variety of dimension $n$ and is properly contained in $\mathbb{A}^{n+1}$, it itself must be of dimension $n$, and so is the zero set of a single polynomial $F \in k[x_1, ..., x_{n+1}]$. The function $f$ then assigns a point $p \in \mathbb{A}^n$ to the necessarily unique zero of $F(p)$, viewed as a polynomial in $x_{n+1}$. If we Noether normalize, then we may assume $F$ is monic in $x_{n+1}$. So, write $F(p) = (x_{n+1} - f(p))^m$. Since the coefficient of the $(m-1)^{\text{th}}$ power of $x_{n+1}$ in $F$ is a polynomial equal to $-m \cdot f(p)$ for all $p$, and we are in characteristic $0$, we divide to obtain $f$ is a polynomial function as desired.
I'm thinking there ought to be a more elegant proof of this even when $k$ is not algebraically closed or of nonzero characteristic, but I can't seem to figure it out.