= Solution
For <algebraic varieties> that are <irreducible topological spaces>, a <rational map> $X\dashrightarrow Y$ is an equivalence class of <morphisms> $U\to Y$ on nonempty open <subsets> of $X$, with two representatives equivalent if they agree on a nonempty open <intersection>. Such <subsets> are dense. \b[A <birational map> has a rational inverse]; equivalently it restricts to an <isomorphism> between dense open <subsets>. In <function field> language it induces an <isomorphism> $k(Y)\cong k(X)$ over the ground <field>.
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