TITLE: True or False: $f$ is injective if and only if $f^*$ is surjective where $f^*$ is corresponding to the pullback to $f$.
QUESTION [5 upvotes]: Let $f: X\rightarrow Y$ be a morphism of affine varieties and $f^*: A(Y)\rightarrow A(X)$ the corresponding homomorphism of the coordinate rings. The question is whether this is true or false:
$f$ is injective if and  only if $f^*$ is surjective. 
The "only if" part is false. Here is a counterexample:
$$X=\mathbb{A^1}, Y=V(x^2-y^3)\\
f: X\rightarrow Y, t \rightarrow (t^3,t^2)\\
f^*: A(Y)\rightarrow A(X), (\bar{x},\bar{y})\rightarrow (t^3,t^2)$$
In this example $f$ is bijective, but $f^*$ is not surjective, since it does not map anything to $t$.
I cannot prove the "if" part or construct a counterexample of it. Thanks for any help!

REPLY [4 votes]: Assume $f^\ast$ is surjective and let us prove that $f$ is injective. Let $x_1,x_2\in X$ be two points with $y:=f(x_1)=f(x_2)$. Assume that $x_1\ne x_2$, then there is some function (the restriction of some coordinate function, say) $\varphi\in A(X)$ with $\varphi(x_1)\ne\varphi(x_2)$. However, $f^\ast$ is surjective, so there is some $\psi\in A(Y)$ with $f^\ast(\psi)=\varphi$. This implies 
$$\psi(y)=\psi(f(x_1))=f^\ast(\psi)(x_1)=\varphi(x_1)\ne\varphi(x_2)=f^\ast(\psi)(x_2)=\psi(f(x_2))=\psi(y),$$
which is a blatant contradiction. We must have had $x_1=x_2$.