Unwinding confusion about (bi)rational maps

I am often confused with birational maps, or just rational maps in general, because they are not quite maps in the usual sense. Let me try to undo my confusion by rewriting some materials on these confusing objects in this posting. I am following Vakil's online book.

Given schemes $X, Y,$ we define a rational map $X \dashrightarrow Y$ to be an equivalence class of scheme maps $\alpha : U \rightarrow Y,$ where $U \subset X$ is a dense open subset with the following equivalence relation: $(\alpha, U) \sim (\beta, V)$ means there is a dense open subset $W \subset U \cap V$ such that $\alpha|_{W} = \beta|_{W}.$

Remark. We often assume that $X$ is reduced because then dense open subsets are precisely scheme-theoretically dense open subschemes of $X,$ the notion that works better than the usual (topological) density in various situations. Using the latter notion gives an analogous notion to that of rational maps on $X,$ which seems to be studied by some people.

A rational map $X \dashrightarrow Y$ is said to be dominant if there is any representative $U \rightarrow Y$ whose image is dense in $Y.$ A dominant rational map $\pi : X \dashrightarrow Y$ is called birational if there is another dominant rational map $\phi : Y \dashrightarrow X$ such that

• $\pi \circ \phi \sim \mathrm{id}_{Y}$ and
• $\phi \circ \pi \sim \mathrm{id}_{X},$

on some dense open subsets (of $X$ and $Y,$ respectively).

When $X$ and $Y$ are reduced, this complicated looking notion means some thing simple: that is $X$ and $Y$ are birational if and only if there are some isomorphic dense open subsets $U \subset X$ and $V \subset Y.$

Let $k$ be a field. Then integral finite type $k$-schemes with dominant rational maps form a category. If we have a morphism $X \dashrightarrow Y,$ it sends the generic point of $X$ to that of $Y,$ so we get a corresponding function field extension $K(Y) \rightarrow K(X).$ It turns out that this establishes an equivalence between the category of integral finite type $k$-schemes with dominant rational maps over $k$ and the category of field extensions that are finitely generated over $K.$ (See Proposition 6.5.7 and 6.5.D in Vakil.)

Separatedness makes rational maps easier. If the target of a rational map is separated (over some base scheme), then it is easier to study the rational map, essentially due to the following fact (from 10.2.2. of Vakil):

Theorem (Reduced-to-Separated). Let $X, Y$ be schemes over a scheme $S.$ Suppose that $X$ is reduced and $Y$ is separated over $S.$ If two $S$-scheme maps from $X$ to $Y$ agree on a dense open subset of $X,$ then they must be identical.

How does this help us when studying rational maps? Let $X$ be reduced, defined over a field $k,$ and $Y$ separated over $k.$ A rational map $X \dashrightarrow Y$ defined over $k$ is giving a $k$-scheme map $U \rightarrow Y,$ where $U \subset X$ is a dense open subset. If there is another representative $V \rightarrow Y,$ the above theorem now tells us that the two representatives agree on $U \cap V.$ This implies that we can glue these two maps to get another representative $U \cup V \rightarrow Y.$ Note that we can glue arbitrarily many representatives this way, and get a scheme map $W \rightarrow Y,$ where $W$ is the union of all dense open subsets coming from the representatives. This $W$ is hence unique and deserves a name. It is called the domain of definition of the rational map $X \dashrightarrow Y.$

This is great because in this situation (where the source is reduced and the target is separated), we can get an actual map that contains the information about a rational map. This is much less confusing!