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I'm trying to understand the proof of the following theorem on Page 17 of Guillemin and Pollack's Differential Topology:

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

The proof proceeds with two main steps:

  1. Showing that the image of any open set in $X$ is an open set in $f(X)$ (this apparently proves that $f(X)$ is a manifold), and then

  2. Checking that $f : X \rightarrow f(X)$ is a diffeomorphism.

I understand the the argument that $f$ maps open sets to open sets (which is argued by contradiction considering a sequence), but I don't understand why this guarantees that $f(X)$ is a manifold.

Any hints/explanations would be greatly appreciated!

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1 Answer 1

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You can build the charts on $f(X)$ explicitly. Let $q\in f(X)$ and consider $p=f^{-1}(q)$. Let $(U,\varphi:U\to\mathbb{R}^n)$ be a chart on $X$ centered at $p$. It follows that $V=f(U)\subset f(X)$ is an open neighborhood of $q$, moreover, we see

$$ (V,(\varphi\circ f\vert_V^{-1}):V\to\mathbb{R}^n) $$

defines a chart on $f(X)$ centered at $q$. Now just check the transition maps are smooth (this follows since they are smooth on $X$).

Now it is clear that $f$ is a diffeomorphism onto its image because 1) $f$ is bijective onto its image by assumption and 2) $f$ is a local diffeomorphism by the above.

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  • $\begingroup$ Thank you! That cleared up my confusion perfectly $\endgroup$ Commented Dec 22, 2023 at 18:49

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