Two affine varieties (defined as irreducible algebraic sets) are isomorphic if there are mutualy inverse polynomial maps between them. Is it true that if I have an affine variety $V$ and an invertible polynomial map $f:V\rightarrow S$ to an algebraic set $S$, then $S$ must also be an affine variety?
This seems to be true to me because such polynomial maps should induce an isomorphism on the coordinate rings of $V$ and $S$, so if $\Gamma (V)$ is a domain, then so is $\Gamma (S)$.
I guess I still don't have a concrete intuition for why this should be true though. For example, if we let $V=(t, t^2)$ and $S=(t, t^2, t+t^6)$, then $V = V(y-x^2)$ which is irreducible and $S=V(y-x^2, z-x-y^3)$. We can let $f_3(x, y)=x+y^3$ to construct a polynomial map between $V$ and $S$, and then projection to go back from $S$ to $V$.
Now if the above argument is true I feel like it should be clear why $I(S)$ should be prime, but I am having trouble working through the details in this concrete example.