Pullback of F is injective if and only if the image of F is dense - proof check This has been asked here and here before, but not to my satisfaction - the first answer is not incredibly detailed to my mind, neither is the second - so hopefully this will be allowed as a question on its own. Specifically, I have tried to write out in detail the proof, and am looking for feedback on improvement/where I've totally misunderstood something. The question is exercise 2.5.1 of Smith's Invitation to Algebraic Geometry. 
I'll use $ F: V \to W $ to denote a morphism of affine algebraic varieties, and $ F^* : \mathbb{C}[W] \to \mathbb{C}[V] $ the pullback of $F$ on the coordinate rings. The question: show that $F^*$ is injective if and only if $F(V)$ is dense in $W$. To do so we need to use the characterisation of density as having nonempty intersection with all open sets of $W$. In both directions I prove it by contradiction, which is (kind of) gross.
Assume $F^*$ is not injective, so that there exists $ g \in \mathbb{C}[W]$ such that $g \neq 0$ on $W$ and $ (gF)(x) = 0 $ for all $ x \in \mathbb{C}[V]$. We seek a contradiction of the second statement. Consider the open set $ U = \{y \in W : g(y) \neq 0\}$. If $ y \in U \cap F(V)$, then there exists $ x \in V $ so that $y = F(x)$, and $ g(y) \neq 0$. Thus $(gF)(x) \neq 0$, a contradiction.
Now assume $F(V)$ is not dense, i.e. there exists some nonempty open set $U \subseteq W$ so that $U\cap F(V) = \emptyset$. So there exists some family of polynomials $p_i$ with $i \in I$ so that $U = \{y \in W : p_i(y) \neq 0, i \in I\}$. Hence $U \cap F(V) = \emptyset$ implies that $(p_i F)(x) = 0$ for all $ i \in I$ and $x \in V$, which by the injectivity of $F$ means $p_i = 0$, i.e. $U = \emptyset$, a contradiction. End of proof.
I'm fairly confident in my writeup, but I'm also not sure if there might be better (maybe easier) ways of arguing the proof. I would like to avoid (if possible) more machinery than necessary, since at this point in time in the text Smith has not (formally) introduced many concepts. Any critique is welcome!
 A: Your proofs are correct. There are a few minor improvements we could consider making.
For the first proof, we can rephrase it to avoid it being by contradiction without too much trouble. Consider the following:

Suppose $F(V)\subset W$ is dense. For any nonzero function $g\in k[W]$, we consider $F(V)\cap D(g)$ (where $D(g)$ is the open set where $g$ doesn't vanish). By the definition of density, this is nonempty. Pick a point $y\in F(V)\cap D(g)$. Then $(F^*(g))(y)\neq 0$, so $F^*(g)$ cannot be zero. Thus $\ker F^*=0$ and the map $F^*$ is injective.

In the second proof, the idea is good but the phrasing with the $p_i$ bugs me a little. Here's how I'd rephrase it to be a little more straightforwards while keeping the same idea.

Suppose $F(V)\subset W$ is not dense. Then there exists an open subset $U$ so that $F(V)\cap U = \emptyset$, which is equivalent to $F(V)\subset U^c$. As $U^c$ is a proper closed subset of $W$, it is contained in a set of the form $V(p)$ for $p\in k[W]$ a nonzero nonunit. Then $F^*(p)=0$, so $F^*$ is not injective.

If we wanted to run the proof in the forwards direction instead, here's how we might want to do that:

Suppose $F^*:k[W]\to k[V]$ is injective. Then for any nonzero $p\in k[W]$, we see that $F^*(p)\neq 0$. This means that there's a point $v\in V$ so that $(F^*(p))(v)\neq 0$. In turn, this implies that $F(V)$ is not contained in any set of the form $V(p)$ for $p\neq 0$. As every nontrivial open subset of $W$ contains a set of the form $D(p)$ for $p\neq 0$, this implies that $F(V)$ intersects every nonempty open subset of $W$ and is therefore dense.

Personally, I like the first version of part 2 better, but the second version can also be instructive.
