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Clearly I suppose to put the condition $n > 1$ in use. So my proof must went wrong.. Could someone help me take a look at it? Thanks!

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.

By local immersion theorem,

Local Immersion Theorem Suppose that $f: X \to Y$ is an immersion at $x$, and $y = f(x)$. Then there exist local coordinates around $x$ and $y$ such that $$f(x_1, \dots, x_k) = (x_1, \dots, x_k, 0, \dots, 0).$$ In other words,$x$ is locally equivalent to the canonical immersion near $x$.

Therefore, the image of $f$ is open, since $(x_1, \dots, x_k)$ is open in $X$ so $(x_1, \dots, x_k, 0, \dots, 0)$ is open in $Y$. Also, because $X$ is compact, $f$ is proper. Therefore, $f$ is an embedding. According to the theorem, $f$ is an embedding.

Theorem An embedding $f: X \to Y$ maps $X$ diffeomorphically onto a submanifold of $Y$.

However, continuous functions preserves compactness, hence the image of $f$ is closed. The only closed and open subset of $Y$ is $Y$ itself. Hence $f$ is surjective. Therefore, since $f$ maps $X$ diffeomorphically onto a submanifold of $Y$, the submanifold is $Y$. And consequently, $f$ is a diffeomorphism.

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  • $\begingroup$ An embedding is also injective. $z \mapsto z^2$ is an immersion $S^1 \to S^1$ that is not injective. $\endgroup$ – Daniel Fischer Aug 19 '13 at 17:50
  • $\begingroup$ @DanielFischer Sorry! I accidentally hit enter before finishing my response. $\endgroup$ – Avi Steiner Aug 19 '13 at 17:57
  • $\begingroup$ Hi @DanielFischer, thanks - I'm sorry I missed the last paragraph. I just added. $\endgroup$ – 1LiterTears Aug 19 '13 at 17:58
  • $\begingroup$ @AviSteiner, I see what you meant, I'm sorry. Not sure how come I left the last paragraph.. Just added. Thanks!! $\endgroup$ – 1LiterTears Aug 19 '13 at 17:58
  • $\begingroup$ "Therefore, since $f$ maps $X$ diffeomorphically onto a submanifold of $Y$", you have not shown that, and - see above - you can't deduce that for dimension $1$. Question: do you know what a covering is? $\endgroup$ – Daniel Fischer Aug 19 '13 at 18:00
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What you have is in fact a "local embedding" instead of a global embedding. You need to be slightly careful since different books may use different definition of "embedding" and "immersion". But anyway, now you have an immersion $f$ which is locally a diffeomorphism

$$M\overset{f}{\to} S^n$$

For a local embedding to fail to be a global one, you must have overlapped image, or equivalently have $f$ failing to be injective. This amounts to saying that $f$ must be a covering map of $S^n$, or equivalently $M$ is a covering space of $S^n$. This rules out the possibility of $S^1$ since its fundamental group is not trivial. Now for $n>1$, since it is well known that they have trivial fundamental group, therefore $f$ must be a 1-folded covering map, or equivalently a global embedding.

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  • $\begingroup$ Aha, got it! Thank you so much shallpion! $\endgroup$ – 1LiterTears Aug 19 '13 at 18:03
  • $\begingroup$ So the theorem says, given a local embedding, I have a local diffeomorphism; given a glocal embedding, I have a glocal diffeomorphism? $\endgroup$ – 1LiterTears Aug 19 '13 at 18:14
  • $\begingroup$ I think you are right. $\endgroup$ – Peng Aug 19 '13 at 18:16
  • $\begingroup$ +1 from me but I just wanted to add that the fact that $f$ is a covering map depends on the fact that $M$ is a compact, boundaryless manifold. There are plenty of non-compact or compact-with-boundary manifolds $M$ which admit immersions into $S^n$ but fail to be diffeomorphisms. For example, if $M$ is the open or closed northern hemisphere and $f$ is the inclusion map. $\endgroup$ – Jason DeVito Aug 20 '13 at 1:29

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