Lee Smooth Manifolds - Lemma (6.14) for Whitney Embedding Theorem

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:

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.

• I think I understand it now, thanks! Just out of curiosity about the incomplete sentence at the end - by the second part, is it that we can iterate and find properly embedded submanifolds of $\mathbb{R}^k$ for every $k = \{2n + 1, \cdots, N-1\}$? Also, the stopping condition is when $N \leq 2n + 1$ so we can't find an injective immersion of M into $\mathbb{R}^{N-1}$, right? Commented May 13, 2022 at 0:17
• Looks like I was distracted. That’s the correct idea, yes. However, it is possible that injective proper immersions into $\mathbb{R}^N$ with $N <2n+1$ exist (see “strong Whitney” if I recall correctly). The point however is that this proof fails to construct them. Anyway, I edited. Commented May 13, 2022 at 6:17