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Let $X$ be a nodal curve, and let $\phi\in Aut(X)$. Let $\tilde{X}$ its normalisation.

Is there a condition on $\phi$ to be lift to $\tilde{X}$?

How about involutions?

Thanks

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    $\begingroup$ $\phi$ induces a map $\phi \circ \nu : \widetilde X \to X$, where $\nu:\widetilde X\to X$ is the normalization. By universal property of the normalization of a variety, this map lifts to a map $\widetilde \phi : \widetilde X \to \widetilde X$. Now if you do this construction with $\phi^{-1}$, you find a lift $\widetilde {\phi^{-1}}$ such that $\widetilde \phi \circ \widetilde{\phi^{-1}}$ is the identity map of $\widetilde X$ on a dense open set, thus everywhere. Therefore $\widetilde \phi $ is an automorphism of $\widetilde X$. $\endgroup$
    – Henri
    Dec 5, 2016 at 22:13

1 Answer 1

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Normalization is a functor. This means that automorphisms, as well as involutions, lift canonically.

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