Problem with compacts in Lemma before Whitney Embedding Theorem

Question

I've been studying Lee's book Introduction to Smooth Manifolds and I'm struggling to understand the last paragraph of the Lemma 6.14 that comes right before Whitney Embedding Theorem. We want to show that if we have a proper embedding of $M$ into $\mathbb{R}^N$, it is possible to embed $M$ properly into $\mathbb{R}^{N-1}$. My problem is when he proves that he projection, indeed I had to proceed slightly different to get until this point. I start with the fact that we have a injective proper immersion from $\mathbb{R}^N$ to $\mathbb{R}^N$ that sends the space in the unitary ball centered in the origin, then we can consider that is $M$ is embedded in $\mathbb{R}^N$, then it is embedded properly and in such a way that it is contained in the unitary ball centered in the origin. I can use the same Lemma 6.13 (there is a vector such that the projection along it is a an injective immersion) but then a problem, at least to me, arises. I have to show that the projection is proper but imagine that we have a manifold that as a subset of $\mathbb{R}^N$ is not compact/closed, how can I guarantee that the intersection $\pi_v^{-1}(K) \cap M$ is compact? I know that the preimage is compact, but not sure about the intersection.

To have a clearer example, imagine $M$ as a section of the open unitary ball, how does the compactness would be assured? Maybe I'm missing an elementary fact of topology, but for now I can't remember.

I really appreciate you help and attention.

Answer 1

Hare's how to verify that $\pi_v^{-1}(K)\cap M$ is compact:

• $\pi_v^{-1}(K)\cap M$ is bounded (because it lies in the intersection of two tubes with non-parallel axes).
• $M$ is closed (because it's properly embedded).
• $\pi_v^{-1}(K)$ is closed (because it's the preimage of a closed set under a continuous map).
• $\pi_v^{-1}(K)\cap M$ is closed (because it's the intersection of two closed sets).
• $\pi_v^{-1}(K)\cap M$ is compact (because it's closed and bounded).