# Immersion is a diffeomorphism

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.

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