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