I am basically asking for a converse of this question. Is it true that if $\phi: V \rightarrow W$ is a closed injective morphism of affine varieties, then the induced $k$ algebra morphism from regular functions on $W$ to regular functions on $V$ is surjective?
Since $\phi$ is closed, $\phi(V)\subseteq W$ is a variety. Intuitively, for $f$ regular on $V$ one would define $g(\phi(x)) = f(x)$ on $\phi(V)$. But is there anything that guarantees that this can be extended to a regular function on $W$?
I know that the categories of affine varieties and of finitely generated $K$-algebras without nilpotents are equivalent, but I've seen here that an epimorphism in the category of $K$-algebras is not a surjective morphism; I would also prefer a category-free approach if it's possible, as I'm not very comfortable with how equivalence of categories behaves w.r.t epimorphisms and monomorphisms.
Best Answer
I think the answer is no. You can consider $C \subset \mathbb{A}^2$ given by $x^2 = y^3$ and the morphism $\phi : \mathbb{A}^1 \to C$ given by $z \mapsto (z^3 , z^2)$. I think that this $\phi$ is a bijection, and it is closed (since closed subsets of curves are simply cofinite subsets), but it is not a closed immersion.