Equivalent conditions to a rational map $f$ being birational

I'm going to copy out a chunk of Hulek's 'Elementary Algebraic Geometry' (page 83).

Let $V$ and $W$ be irreducible quasi-projective varieties.

Theorem 2.49

For a rational map $f : V \to W$ the following statements are equivalent:

(1) $f$ is birational

(2) $f$ is dominant and $f^* : k(W) \to k(V)$ is an isomorphism

(3) There are open sets $V_0 \subset V$ and $W_0 \subset W$ such that the restriction $f\big|_{V_0} : V_0 \to W_0$ is an isomorphism

Proof. The equivalence of (1) and (2) is proved in the same way as in Theorem 1.80.

To show that (3) implies (1), note that in this case $f\big|_{V_0} : V_0 \to W_0$ has inverse $g: W_0 \to V_0$, and by definition $g : W \to V$ is a rational map. Then $g \circ f : V \to V$ and $f \circ g: W \to W$ are rational maps which are the identity maps on $V_0$ and $W_0$ respectively. Since $V_0$ and $W_0$ are dense, it follows that $g \circ f = \mathrm{id}_V$ and $f \circ g = \mathrm{id}_W$.

(...)

I have two problems with the above.

i) I can't see why a birational map must be dominant.

ii) I don't understand the final part of the proof that (3) implies (1). We have that $g \circ f = \mathrm{id}_{V_0}$ and $f \circ g = \mathrm{id}_{W_0}$. Why do we need density to conclude that $g \circ f = \mathrm{id}_V$ and $f \circ g = \mathrm{id}_W$?

Would I be correct in saying that a dominant rational map $h : V \to W$ gives rise to a $k$-homomorphism $h^* : k(W) \to k(V)$, and imposing that $h$ is birational makes $h^*$ an isomorphism?

Thanks

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Only dominant rational maps can give rise to a homomorphism between the function fields. It's worth thinking about this. –  Zhen Lin Mar 14 '12 at 22:56
@ZhenLin I understand this (a rational map $f : V \to W$ being dominant is equivalent to always being able to compose $f$ with a rational map $g : W \to k$; so we can always define $f^*$). I'm not sure that answers my question, or at least not obviously. Why does "$f$ is birational" imply "$f$ is dominant and $f^* : k(W) \to k(V)$ is an isomorphism"? –  Matt Mar 14 '12 at 23:05
Hmmm. Well, perhaps you would prefer a more geometric argument. Suppose $f \circ g$ is a dominant rational map. Do you agree that $f$ must be dominant? Now suppose $f$ is birational and $g$ is its inverse... –  Zhen Lin Mar 14 '12 at 23:17
@ZhenLin OK, I've got it provided I know that $\mathrm{dom}(g)$ is open in $W$. I know this is true in the affine case, so I'll have a think why it's true more generally –  Matt Mar 14 '12 at 23:28
@ZhenLin I'm guessing the definition of $\mathrm{dom}(g)$ where $g$ is a rational map of projective varieties is the union of the domains of all the representations of $g$. In which case, I can see why it's open in $W$. –  Matt Mar 15 '12 at 0:03