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.

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    $\begingroup$ Pick a name for the immersion to start, $f:X \to S^n$. What properties of $f$ can you deduce first? And next? And next? For example, is $f$ surjective? ... $\endgroup$ – Lee Mosher Jun 21 '13 at 18:20
  • $\begingroup$ To start with - $df_x$ is injective $\forall x \in X$. $\endgroup$ – 1LiterTears Jun 21 '13 at 18:25
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    $\begingroup$ But you need some global properties of $f$. Hence my question of whether $f$ is surjective. $\endgroup$ – Lee Mosher Jun 21 '13 at 18:52
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    $\begingroup$ Pay attention to the fact, Jellyfish, that $\dim X = \dim S^n = n$. So what do you know immediately if $df_x$ is injective? $\endgroup$ – Ted Shifrin Jun 21 '13 at 20:13
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    $\begingroup$ So $f$ is a local diffeomorphism? $\endgroup$ – Ted Shifrin Jun 21 '13 at 21:34

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.

  • $\begingroup$ Thank you so much Jack Lee! $\endgroup$ – 1LiterTears Jun 22 '13 at 20:57
  • $\begingroup$ Thanks again Professor Lee! I am really elated getting your help! I just got your book Introduction to Topological Manifolds, and I am very excited to read it! I am also reading your other book, Riemannian manifolds this summer. Could you give me some suggestion reading your three books - for example, in which order should I read them (or concurrently)? Thank you very much. $\endgroup$ – 1LiterTears Jul 2 '13 at 16:18
  • $\begingroup$ @Jellyfish -- they're meant to be read in this order: (1) Introduction to Topological Manifolds, (2) Introduction to Smooth Manifolds, and (3) Riemannian Manifolds. Of course, hardly anybody really reads all three from cover to cover. If you prefer, you can start at the beginning, skip through and focus on the parts that most interest you, and then come back and fill in details when you need them. $\endgroup$ – Jack Lee Jul 2 '13 at 17:41

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