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If $f: S^n \to S^n$ be a local diffeomorphism and $n > 2$, then $f$ is a global diffeomorphism.

I do not know why this should be true. I tried thinking of the $n = 1$ case, $z^2$ is a local diffeomorphism but not a global diffeomorphism . But I failed to deduce from this what properties of $n > 2$ dimensional spheres have different from $S^1$ play a key role in this.

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    $\begingroup$ The case $n=1$ is special, because that is the only case in which $\mathbb{S}^n$ is not simply connected. $\endgroup$
    – Laz
    May 23, 2019 at 20:36

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Since $\mathbb{S}^n$ is Hausdorff, compact and connected, any local homeomorphism $f:\mathbb{S}^n\rightarrow \mathbb{S}^n$ is a covering map (for example, convince yourself with this When is a local homeomorphism a covering map?).
But since $\mathbb{S}^n$ is simply connected for $n\geq 2$, and this object (the Universal covering map) is unique up to isomorphism of covering maps, $f$ is homeomorphic to the trivial cover $\mathbb{Id}_{\mathbb{S}^n}: \mathbb{S}^n\rightarrow \mathbb{S}^n$. This means that their fibers have the same cardinality 1. So, f is a diffeomorphism.
$\textbf{Added}:$ Observe that I never used the smoothness. So that, more generally, any local homeomorphism $f:\mathbb{S}^n\rightarrow \mathbb{S}^n$ is a homeomorphism if $n\geq 2$.

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  • $\begingroup$ Does this argument just show that $f$ has to be a homemorphism? How do we know that the inverse is smooth? $\endgroup$
    – penny
    May 23, 2019 at 21:25
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    $\begingroup$ Penny, smoothness in a local condition: you verify it in a neighborhod of each point using charts, and locally your map is a diffeo, menaning that its inverse is also smooth locally. So globally its inverse will also be smooth. $\endgroup$
    – Laz
    May 23, 2019 at 21:27
  • $\begingroup$ I see, that is very neat! Thank you very much! $\endgroup$
    – penny
    May 23, 2019 at 21:29
  • $\begingroup$ I am glad I could help you. $\endgroup$
    – Laz
    May 23, 2019 at 21:30

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