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Suppose we take an algebraic variety $X$ over $\mathbb{C}$ (I assume reduced). Is the normalization $$\pi:\tilde{X} \to X$$ always bijective on the smooth points of $X$?

Karsten
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    It is an isomorphism over the smooth locus of $X$, as you can see by the local definition in Hartshorne. There should be a slick argument applying the universal property of normalisation to the inclusion of the smooth locus $X_{sm} \rightarrow X$, but I don't see it yet. –  Apr 10 '14 at 15:15

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Yes!
The key point is that the smooth points of $X$ are regular and a fortiori normal.
Now the set of normal points of a variety is open [ if you want to show-off tell your rival that this is a property valid for all quasi-excellent schemes :-)] and since normalization does not change any normal open subset of $X$, you can conclude that normalization is not only bijective but even an isomorphism over the smooth points of $X$.

Georges Elencwajg
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