Pullback $\Bbb C [W] \xrightarrow{F^\sharp} \Bbb C[V]$ is surjective iff $F: V \to W$ defines an isomorphism between $V$ and some subvariety of $W$. 
Show that the pullback $\Bbb C [W] \xrightarrow{F^\sharp} \Bbb C[V]$ is surjective if and only if $F: V \to W$ defines an isomorphism between $V$ and some algebraic subvariety of $W$.

This problem originates from the book "An invitation to algebraic geometry". My work is the following.
We assume that $F^\sharp : \Bbb C [W] \to \Bbb C [V]$ is surjective. Since the coordinate rings are $\Bbb C$-algebras and $F^\sharp$ is an algebra homomorphism by the first isomorphism theorem we have that $\Bbb C[W]/ \ker F^\sharp \cong \Bbb C[V]$.
Now also $\ker F^\sharp = \{ P \in \Bbb C[W] : F^\sharp(P)=0 \}$, but $F^\sharp (P)=0 \iff P(F(x)) = 0$ for every $x \in V$. This implies that $P \in I(F(V))$ that is $P$ belongs to the polynomials that vanish at $F(V)$. So I have that $\ker F^\sharp = I(F(V))$ and $$\Bbb C[W]/ \ker F^\sharp  = \Bbb C[W]/ I(F(V)) \cong \Bbb C[V]$$
But I still don't have the isomorphism $V \cong W'$ for $W' \subset W$. I think that $W' = F(V)$, but I cannot prove this?
 A: This is true for any algebraically closed field $k$, so I will use $k$ to denote an algebraically closed field.

First Direction
Suppose that $F:V \to W$ is an isomorphism onto a subvariety $X$ of $W$. Then we have maps
$$
V \to X \hookrightarrow W,
$$
where the first map is an isomorphism. Taking pullbacks, we get
$$
k[V] \leftarrow k[X] \leftarrow k[W].
$$
The map $k[X] \to k[V]$ is an isomorphism since $F:V \to X$ is an isomorphism. Since $X$ is a subvariety of $W$, the map $k[W]\to k[X]$ is surjective (to see this, take any polynomial function $F:X \to \mathbb{A}^1$ and extend it to $W \to \mathbb{A}^1$ by using the same polynomial). In particular, the map $k[W] \to k[V]$ is a composition of surjective maps, so it is surjective.

Second Direction
Suppose conversely that $k[W] \to k[V]$ is surjective. Let $F(V)$ be the set-theoretic image of $V$ in $W$ and let $\overline{F(V)}$ be its Zariski closure, so that $\overline{F(V)}$ is a subvariety of $W$. Then we have maps
$$
V \to \overline{F(V)} \hookrightarrow W,
$$
and taking pullbacks gives
$$
k[V] \leftarrow k[\overline{F(V)}] \leftarrow k[W].
$$
By definition of $\overline{F(V)}$, the map $V \to \overline{F(V)}$ is dominant, which means that the map $k[\overline{F(V)}] \to k[V]$ is injective (it is a standard result that a morphism of affine varieties is dominant if and only if its pullback is injective. If you don't know that fact, you should prove it as an exercise). Since $k[W] \to k[V]$ is surjective, the map $k[\overline{F(V)}] \to k[V]$ is also surjective. We have shown that
$$
k[\overline{F(V)}] \to k[V]
$$
is injective and surjective, so it is an isomorphism. It follows that $V \to \overline{F(V)}$ is an isomorphism, so we are done because $F$ is an isomorphism onto the subvariety $\overline{F(V)}$ of $W$.

More algebraic method for second direction
Let $V\subseteq \mathbb{A}^m$ and $W \subseteq \mathbb{A}^n$ be affine varieties over an algebraically closed field $k$, and let $F:V \to W$ be a morphism of affine varieties. Suppose that the pullback $F^*:k[W] \to k[V]$ is surjective. We have
$$
k[W] = k[x_1,\ldots, x_n]/\mathcal{I}(W)
$$
by definition of coordinate rings. Then $\ker F^*$ is an ideal of $k[W]$, so it equals $I/\mathcal{I}(W)$ for a unique ideal $I$ of $k[x_1,\ldots, x_n]$ containing $\mathcal{I}(W)$. Define $Z = \mathbb{V}(I)\subseteq \mathbb{A}^n$. Then $Z$ is an affine variety, and we have $Z \subseteq W$ since $I \supseteq \mathcal{I}(W)$. We claim that $F$ is an isomorphism onto $Z$.
Claim: We have $F(V)\subseteq Z$.
Proof: Let $a \in V$, so that $F(a) \in F(V)$ and let $f \in I$. Then $f(F(a)) = (F^*[f])(a) = 0(a) = 0$, where I have written $[f]$ to denote the element $f + \mathcal{I}(W)$ of $I/\mathcal{I}(W) = \ker F^*$. In particular, $f(b) = 0$ for all $f \in I$ and $b \in F(V)$, so $F(V)\subseteq \mathcal{V}(I) = Z$. $$\tag*{$\blacksquare$}$$
Now, we have a morphism $G:V \to Z$ of affine varieties, such that the composition $V \overset{G}{\to} Z \overset{i}{\hookrightarrow} W$ is equal to $F$. Taking pullbacks, we see that the composition $k[W] \overset{i^*}{\to} k[Z] \overset{G^*}{\to} k[V]$ is equal to $F^*$, which is surjective, and therefore $G^*$ is surjective.
Claim: The map $G^*:k[Z] \to k[W]$ is also injective.
Proof: Suppose that $f + I \in \ker G^*$, where $f \in k[x_1,\ldots, x_n]$. Since $G^*f$ is the map $f \circ G:V\to \mathbb{A}^1$, we have $f(G(a)) = 0$ for all $a \in V$. But for $a \in V$, we have $G(a) = F(a)$ by definition of $G$, which means that $f(F(a)) = 0$ for all $a \in V$, and therefore $F^*f:V \to \mathbb{A}^1$ is zero, which means that the element $f + \mathcal{I}(W)$ of $k[W]$ is in the kernel of $F^*$, and therefore $f \in I$ by definition of $I$. It follows that $\ker G^* = 0$, so $G^*$ is injective. $$\tag*{$\blacksquare$}$$
We have shown that $G^*$ is surjective and injective, which means that it is an isomorphism. Therefore, the morphism $G:V \to Z$ is also an isomorphism, so we are done.
