Suppose $X$ is a smooth, compact, connected $n$-manifold without boundary which admits an immersion to $S^n$. Show that if $n>1$, then this immersion is a diffeomorphism.
Thanks for the very inspiring mentors, here I got some thoughts
$df_x$ is bijective. Because the tangent planes of the domain has the same dimension as domain; the tangent space of the codomain has the same dimension as codomain. But dim$X = n$, dim$S^n = n$, so $df_x$ maps from dim$n$ to dim$n$. Given immersion, $df_x$ is injective therefore bijective.
On the other hand, $f$ being an immersion told at that $df_x$ is nonsigular, hence a local diffeomorphism. I got stuck extending local diffeomorphism to global diffeomorphism. Is there a general strategy to achieve this(when this is true)?
Thank you.
Best Answer
Once you know that $f$ is a local diffeomorphism, to conclude that it's a global diffeomorphism you just need to show that it's bijective. Surjectivity is pretty easy: Because $X$ is compact, $f(X)$ is also compact, and because $S^n$ is Hausdorff, $f(X)$ is closed in $S^n$. On the other hand, the fact that $f$ is a local diffeomorphism implies that it's an open map, and thus $f(X)$ is open. Since $S^n$ is connected, $f(X)$ is all of $S^n$.
Injectivity is quite a bit harder. The only proof I know uses the theory of covering spaces. Because $f$ is a proper local homeomorphism, it's a covering map (which is another way to prove surjectivity), and because $S^n$ is simply connected, it follows that $f$ is injective.
One place to read about covering spaces is in my book Introduction to Topological Manifolds.