Suppose that $F: X \to Y$ is a morphism of (affine) varieties $X$ and $Y$. Show that $F'$ is injective if $F(X)$ is dense in $Y$.

Question

Suppose that $F: X \to Y$ is a morphism of (affine) varieties $X$ and $Y$ and $F':\Bbb C[Y] \to \Bbb C[X]$ the pullback. Show that $F'$ is injective if $F(X)$ is dense in $Y$.

Suppose that $F(X)$ is dense in $Y$, then for all open $U \subset Y$ we have $U \cap F(X) \ne \emptyset$. That is $\exists x_0 \in U \cap F(X)$. Now since $x_0 \in F(X)$ we have that $x_0=F(y_0)$ for some $y_0 \in X$.

Then if I assume that $F'$ is not injective I have that for nonzero $g \in \Bbb C[Y] : F'(g)(x)=g(F(x)) =0, \forall x \in X$.

I think that I can get a contradiction from the assumption that $x_0 \in U$ also and that $U$ is the complement of an affine variety and contains points that aren't the zeros of any polynomials, but I don't know how to say this. What should I with this information?

Is it because then $F'(g)(y_0)=g(F(y_0))=g(x_0)$ is for some reason nonzero?

Indirect answer: such morphisms of varieties/schemes are called $\textbf{dominant}$, see Exercise II.2.18(b) from Hartshorne for the general statement. — CZL, Apr 13 '22 at 13:35
As you can see from the duplicate list, this has been addressed several times on MSE before. Please do some searching next time before asking to make sure your question has not already been asked and answered before. — KReiser, Apr 13 '22 at 16:32

Suppose that $F: X \to Y$ is a morphism of (affine) varieties $X$ and $Y$. Show that $F'$ is injective if $F(X)$ is dense in $Y$.

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