I am trying to understand the proof of the following theorem in Differential Topology by Guillemin and Pollack (page 17).

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

The "difficult" part of the proof shows by contradiction that the image of any open set $W$ of $X$ is an open subset of $f(X)$; this implies that $f(X)$ is a manifold. But I do not understand the "trivial" part of the proof in the book:

  • It is now trivial to check that $f:X\to f(X)$ is a diffeomorphism, for we now know $f$ to be a local diffeomorphism from $X$ to $f(X)$.
  • Since it is bijective, the inverse $f^{-1}:f(X)\to X$ is well defined
  • But locally $f^{-1}$ is already known to be smooth.

How do we "now" know that $f$ is a local diffeomorphism from $X$ to $f(X)$ and why "locally $f^{-1}$ is already known to be smooth"?


So we know that $f:X \rightarrow Y$ is an embedding between two smooth manifolds $X$ and $Y$, and that it is at least a local diffeomorphism from $X$ to its image $f(X) \subseteq Y$. Being an embedding, $f$ is injective and furthermore $f$ by definition is surjective onto its image $f(X)$. So there is nothing getting in the way of $f:X \rightarrow f(X)$ actually being a global diffeomorphism.

The last two points highlight that since $f$ is bijective onto $f(X)$, a set-theoretic inverse $f^{-1}:f(X) \rightarrow X$ exists, i.e. we haven't said anything yet about continuity, smoothness, etc. However, since $f$ is a local diffeomorphism, locally $f^{-1}$ is smooth, so now we can say the set-theoretic inverse $f^{-1}$ adopts this smoothness from all the local neighborhoods patched together.

  • $\begingroup$ "... and that it is at least a local diffeomorphism" Why? $\endgroup$ – user6 Feb 26 at 14:05
  • $\begingroup$ In the proof leading up to that point, it is shown that $f(X)$ is a smooth manifold, hence locally the image of $f$ is smooth. It is also injective globally (hence locally) and surjective globally by definition of the image $f(X)$ (hence locally). So locally $f$ is smooth and invertible, and applying the same reasoning to $f^{-1}$ shows it's locally a diffeomorphism. $\endgroup$ – BenCWBrown Feb 26 at 16:03

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