[Math] Is a bijective morphism of quasi-affine smooth varieties an isomorphism

Question

We have a bijective morphism $f:X\to Y$ of quasi-affine varieties (say over $\Bbb{C}$).

Can $f$ fail to be an isomorphism if $X$ and $Y$ are smooth?

Answer 1

Hartshorne gives the Froebenius morphism as an example (exercise I.3.2 b). If $k=\bar k$, and $\operatorname{char}(k)=p\neq 0$, then the map $\Bbb A^1_k\to \Bbb A^1_k$ given by $x\mapsto x^p$ is bijective (and even a homeomorphism), but not an isomorphism of varieties.