 Solution

A rational map from irreducible $X$ to $Y$ is an equivalence class of morphisms $F_U:U\to Y$ defined on nonempty open subsets $U\subseteq X$. Two representatives are equivalent if they agree on a nonempty open subset of their intersection.

Irreducibility makes all these nonempty opens dense and ensures their intersections are nonempty, so the equivalence expresses agreement at general points. A rational map need not extend across all of $X$, and it need not be dominant. An injective homomorphism of function fields additionally requires a dominant morphism of affine varieties, or the corresponding condition on a rational map.