Let $M,N$ be nonempty, smooth manifold of the same dimension. Let $N$ be compact and $M$ be connected. Suppose that $f:N\rightarrow M$ is an embedding. We want to show that $f$ is a diffeomorphism.

Since $f$ is an embedding, then we know that

  1. $f(N)\subset M$ is a smooth submanifold, and

  2. $f:N\rightarrow f(N)$ is a diffeomorphism. That is we identify the domain $N$ with its image $f(N)\subset M$ such that $N\subset M$.

We would like to show that $f$ is a diffeomorphism. So, we need to show $f$ is an invertible smooth map.

  • $\begingroup$ In the first paragraph you mean $f : N \rightarrow M$ ? $\endgroup$ – Si Kucing Nov 29 '17 at 18:49
  • $\begingroup$ Yes, I mean $f:N\rightarrow M$ $\endgroup$ – user394412 Nov 29 '17 at 18:52

I found the answer here: Smooth embeddings that are homeomorphisms but not diffeomorphisms

Since we have an embedding, then $f$ is an immersion.

Since $f$ is an immersion and the dimension of $M$ and $N$ are the same, then we have a local diffeomorphism.

Therefore, $f^{-1}:M\rightarrow N$ is also smooth. We have shown $f$ is an invertible smooth map. Hence, $f$ is a diffeomorphism.

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  • 2
    $\begingroup$ You also need to show that $f$ is surjective. $\endgroup$ – Jack Lee Nov 29 '17 at 23:07
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    $\begingroup$ How would I show that? $\endgroup$ – user394412 Nov 30 '17 at 18:16
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    $\begingroup$ Hint: Connectedness of $M$ is crucial. Do you know the typical argument for showing some subset of a connected space is the whole space? $\endgroup$ – Jack Lee Nov 30 '17 at 18:20
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    $\begingroup$ No I do not know this. $\endgroup$ – user394412 Nov 30 '17 at 18:27
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    $\begingroup$ @SeleneAuckland: Yes, that's the argument I was referring to. $\endgroup$ – Jack Lee Jun 20 '19 at 19:36

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