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:
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
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!
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.