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Let $V,W%$ be irreducible quasi-projective varieties.

A rational map $f : V \to W$ is called birational if there is a rational map $g : W \to V$ with $f \circ g = \mathrm{id}_W$ and $g \circ f = \mathrm{id}_V$.

But $\mathrm{dom}(\mathrm{id}_V) = V$, doesn't it? So we must have that $\mathrm{dom}(f) = V$ and $f(V) = W$, and similarly for $g$. So $f$ and $g$ give an isomorphism between $V$ and $W$. What's going on?


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It's an equality of rational maps. Two rational maps are equal if they agree on the intersection of their domains. – Dylan Moreland Mar 14 '12 at 22:07
@DylanMoreland Didn't know that. Thanks, Dylan – Matt Mar 14 '12 at 22:11
@DylanMoreland So would $f$ and $g$ give an isomorphism between $\mathrm{dom}(f) \cap g(W)$ and $\mathrm{dom}(g) \cap f(V)$? EDIT: Though I can't see why the images of $f$ and $g$ have to be open, so maybe not. – Matt Mar 14 '12 at 22:15

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