$\varphi_n \colon \mathbb{A}^1 \to \mathbb{A}^2$ is the polynomial map given by $X \mapsto (X^2,X^n)$; show that if $n$ is even, the image of $\varphi_n$ is isomorphic to $\mathbb{A}^1$, and $\varphi_n$ is 2-to-1 outside 0.
Miles Reid, Undergraduate Algebraic Geometry, Problem 4.3. (first part)
Showing $\varphi_n$ is 2-to-1 is simple. Clearly $\pm \alpha$ both map to the same element, and these are the only two. If it is 2-to-1, how can it be an isomorphism ? How can there be an inverse (polynomial) map, to $\varphi_n$, when it's not injective?
1) It is false that if $n$ is even the map $\phi_n$ is 2-to-1 outside zero:
That map is injective if the base field $k$ has characteristic $2$ .
2) It is false even in characteristic $\neq 2$ that if $n$ is even the image of $\phi_n$ is isomorphic to $\mathbb A^1(k)$:
For example, if $k=\mathbb F_3$ then $\mathbb A^1(k)$ has three elements whereas the image of $\phi_n$ has only 2 .
3) If $k$ is algebraically closed of characteristic $\neq 2$, the result is indeed true:
The image of $\phi _n$ is the curve $Y=X^{\frac n2}$, which is isomorphic to $\mathbb A^1(k)$ like all graphs of regular functions $\mathbb A^1(k)\to\mathbb A^1(k)$ .