# Why are local diffeomorphisms between spheres are actually diffeomorphism

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.

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

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 $$n\geq 2$$.
• Does this argument just show that $f$ has to be a homemorphism? How do we know that the inverse is smooth? May 23, 2019 at 21:25