The question states that if $X$ is irreducible and quasiprojective, and $U$ is a nonempty open subset of $X$, then the inclusion map is a birational equivalence.
The definition I am given for birational is that the map $f: X \to Y$ is a dominant rational map with a dominant rational inverse (i.e. a $g: Y \to X$ such that $f \circ g = Id_Y$ and $g \circ f = id_X$).
My thoughts so far:
If $X$ is irreducible then we have that $U$ is dense in $X$, and so clearly the image of the inclusion map is dense and we have a dominant rational map $\iota: U \to X$.
I can see that this is the easy part of the problem. Where I am struggling is understanding how we obtain an inverse. From what I can tell the inverse would need to have domain of definition being $U \subset X$ (which I think should come from the left inverse guaranteed by injectivity of $\iota$), which would then give us $\iota \circ \iota^{-1} = Id_U$. But the problem is that I don't see why that should also correspond to the composition being $\iota^{-1} \circ \iota = Id_X$.
I understand that rational maps are defined via equivalence classes wherein two rational maps are the same if they agree on a dense open set, which would mean that $Id_X$ agrees with $\iota^{-1}$ on the dense open $U$, but I don't understand how I would get the composition to be equal to the identity of $X$ the codomain of the inverse is $U$.
We have already proven that a dense open subset $U \subset X$ has a function field isomorphic to $X$, but we do not yet have that the equivalence of definitions $k(X) \cong k(Y)$ iff $X$ is birational to $Y$.
Any help here would be appreciated.
Best Answer
Recall the definition of rational map $f:X \to Y$: it is an equivalence class $(f_U,U)$ of morphisms $f:U \to Y$ where $U \subset X$ is open. We consider $(f_V,V) \sim (f_U,U)$ if they agree agree on $U \cap V$, i.e. that $f_U|V=f_V|U$.
So now consider the inclusion $i:U \to X$ as a rational mapping. Also consider $id_X: X \to X$ as a rational map. These two agree on the open set $U$. So they are equal as rational maps. But the identity is obviously a birational equivalence.
See also my old answer to a similar question here: the definition of "Birational Equivalence"