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For algebraic varieties that are irreducible topological spaces, a rational map $X⇢Y$ 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. 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.

Work over the usual algebraically closed ground field of characteristic zero. Away from $x=0$, a line of slope $t=y/x$ through the singular point meets the curve with $t^2=x-1$. ThusSubstitution verifies the defining equation. On $x\ne 0$ and $t^2+1\ne 0$, the two maps are inverse morphisms, so this is a birational map to ${\mathbb{A}}^1$. The two parameters $t=\pm i$ both map to the node $(0,0)$, explaining why the globally defined parametrization is not a global isomorphism.