Let $X \subseteq k^n, Y \subseteq k^m$ be algebraic varieties. Suppose we have a regular map, $\phi:X \rightarrow Y$ induced from a polynomial. $\phi$ induces a $k$-algebra homomorphism on the coordinate rings,
$$ \phi' : A(Y) \rightarrow A(X), \quad [\eta] \mapsto [\eta \circ \phi ] $$
Where $A(K):= k[x_1, \ldots, x_t]/I(K)$. for an algebraic set $K \subseteq k^t$.
I noticed that if $\phi$ is surjective, then $\phi'$ is injective. Is the converse true?
If $\phi'$ is injective then is $\phi$ surjective?
My thoughts: If statement were true. Prove contrapositive. Then there must exists two maps $f,g \in k[x_1, \ldots, x_n]$ such that
(i) $f(y) \not= g(y)$ (this implies $[f]\not= [g]$ in $A(Y)$)
(ii) $f(\phi(X)) = g(\phi(X))$.
This does not seem plausible in general.
No, it does not imply that it is surjective. For example, the map $A^1 \setminus {0} \to A^1$ by inclusion induces an injection on the coordinate rings. This is because any polynomial is determined by its restriction to $A^1 \setminus 0$.
What is true is that $\phi$ is dominant; i.e. the image is dense. (To prove this, consider the closure of the image. If the image is not dense, then the closure is a proper subvariety. In particular, there is a nonzero polynomial $f$ that is zero on $\overline{im(\phi)}$, and $f$ must be in the kernel of the pullback map.)