Q: Comorphism of a polynomial map I'm reading through this paper right now, and I'm stuck on page 8, exercise 18 c):  

Let $\phi:W\to V$ be a polynomial map of finite dimensional vectorspaces. Define $\phi^*:k[V]\to k[W],h\mapsto h\circ\phi$. Show that $\phi^*$ surjective implies $\phi$ injective.

I tried constructing a polynomial map that $\phi^*$ doesn't map to explicitely, but failed horribly. Also, the wording implies that $\phi$ injective$\implies\phi^*$ surjective is not true, but I can't find an example for that. Any help is appreciated.
Edit: I was on the wrong path. You can show this directly: Let $w\ne w'\in W$. There is an $f\in K[W]$ such that $f(w)=0,f(w')=1$, e.g. $f(t_1,...,t_n)=(t_1-w_1)/(w_1'-w_1)$ (assume w.l.o.g. the first coordinate of $w,w'$ differ). Since $\phi^*$ surjective we find $\psi\in K[V]$ such that $f=\phi^*(\psi)$. then $\psi(\phi(w))=f(w)=1\ne 0=f(w')=\psi(\phi(w'))$. Thus $\phi(w)\ne\phi(w')$ for $w\ne w'$, which is what we wanted.
I am still thinking about the example for the reversal not always being true.

A: The proof is in the OP. A counterexample would be $\phi:\mathbb{R}\to\mathbb{R},x\mapsto x^3$ is clearly injective. Then $\phi^{*}:h\mapsto h\circ(x\mapsto x^3)$ but $id_{\mathbb{R}}$ has no preimage under $\phi^{*}$.
