Surjective map on coordinate rings implies the map is injective

Question

Let $X$ and $Y$ be affine varieties and $f: X \rightarrow Y$ a polynomial map. If the induced map on coordinate rings $K[Y] \rightarrow K[X]$ is surjective why this implies that $f$ is injective?

Answer 1

If $x\neq x'\in X$, then there exists $\phi\in K[X]$ with $\phi(x)=0$ and $\phi(x')=1$.
Let $\phi= f^\ast \psi=\psi\circ f$ for some $\psi \in K[Y] $ (surjectivity hypothesis).
Then $\psi( f(x))=(f^\ast \psi)(x)=\phi(x)=0$ whereas $\psi( f(x'))=(f^\ast \psi)(x')=\phi(x')=1$, and of course the inequality $\psi( f(x))\neq \psi( f(x')) $ forces $f(x)\neq f(x')$.

Answer 2

I'll assume you understand the correspondence between maximal ideals of $K[X]$ and points of $X$. I'll also assume the base field $k$ is algebraically closed. Then, for every point $p$ of $X$, there is a unique $k$-algebra homomorphism $\epsilon_p : K[X] \to k$ corresponding to $p$. We compose this with the pullback map $f^* : K[Y] \to K[X]$ to $\epsilon_{f (p)} : K[Y] \to k$. Now, suppose $f (p) = f (p')$. Then $\epsilon_{f (p)} = \epsilon_{f (p')}$, but then that means $\epsilon_p \circ f^* = \epsilon_{p'} \circ f^*$, and hence, $\epsilon_p = \epsilon_{p'}$ if $f^*$ is surjective. But that implies $p = p'$, so $f$ is injective.