I am studying Lee's Introduction to Smooth Manifolds but I am stuck by a lemma for Whitney's Embedding Theorem.
This lemma is trying to prove that: If a smooth n-manifold admits a smooth embedding $\mathbb{R}^N$ for some $N$, then it admits a proper smooth embedding into $\mathbb{R}^{2n+1}$.
In the first part, Lee showed that if $F: M \rightarrow \mathbb{R}^N$ is an arbitrary smooth embedding, then we can find a proper embedding $\varPsi: M \rightarrow \mathbb{R}^N \times \mathbb{R}$ defined by $\varPsi(p) = (G \circ F(p), f(p))$, where $G: \mathbb{R}^N \rightarrow \mathbb{B}^N$ is a diffeomorphism and $f: M \rightarrow \mathbb{R}$ is a smooth exhaustion function. Additionally the image of $\varPsi$ is contained in the tube $\mathbb{B}^N \times \mathbb{R}$.
My question is about the second part in which he replaces $N+1$ by $N$. Why can we replace $N+1$ by $N$ in the previous result? I can see that he uses the case of $N$ to derive the result for $N+1$, but to use this to get the result of $N$, don't we need another arbitrary embedding into $\mathbb{R}^{N-1}$ first?
Here's the screenshot of the lemma and the proof:
Best Answer
This point is slightly subtle (logically speaking).
The idea is that Lee doesn’t need the exact value of $N$ at this point – it’s redefined as “some integer such that there is a smooth proper embedding of $M \rightarrow \mathbb{R}^{N}$”. You’ll notice that it’s slightly more specific than the original definition (“some integer such that there is a smooth embedding $M \rightarrow \mathbb{R}^N$”). But the conclusion doesn’t depend on $N$, and nor do the hypotheses (the assumption is: “*there exists $N$ such that…”) so it’s harmless to choose another $N$ with better proprieties.
If you want to look at it more formally, Lee’s proof consists of two steps. To formulate them best, let $S$ denote the set of integers $N \geq 1$ such that there exists a smooth proper embedding $M \rightarrow \mathbb{R}^N$.
The first part of the proof shows that if $N$ is an integer such that $M$ embeds smoothly into $\mathbb{R}^N$, then $N+1 \in S$.
The second part of the proof is to show that for any integer $N>2n+1$, if $N \in S$ then $N-1 \in S$.
The conclusion is clear once you have these two pieces: by the first part, $S$ is nonempty; by the second part, the smallest element of $S$ must be at most $2n+1$. QED.