# Show this Map $f:M\rightarrow S^n$ is a Diffeomorphism

Let $M$ be a compact, connected smooth manifold of dimension $n>1$, and $f:M\rightarrow S^n$ a smooth map such that the differential $df_x:T_xM\rightarrow T_{f(x)}S^{n}$ has full rank $= n$ for all $x \in M$.

Show that $f$ is a diffeomorphism.

Where I am at:

-Full rank gives local diffeomorphism, so if I can show that $f$ is bijective then by "Global Rank Theorem", this is diffeomorphism.

• Quite surprising to me, but it seems it really holds :) – Peter Franek Jul 17 '14 at 22:51

That $f$ is a local diffeomorphism implies that it is an open map. Hence $f(M)$ is an open subset of $S^n$. But $M$ is compact, so $f(M)$ is also closed. Since $S^n$ is connected, it follows that $f$ is surjective.
Further, since $df_x$ has full rank everywhere, the fibres $f^{-1}(p)$ are discrete. Hence the fibres are finite. Use that to find that $f$ is a covering map. But the sphere $S^n$ is simply connected for $n > 1$, so it only has trivial coverings. Since $M$ is connected, it must be a single-sheeted covering, that is, a homeomorphism.