Set of critical values is compact [closed]

Question

Let $M\subseteq \mathbb{R}^n$ be a compact manifold with $\partial M=\emptyset$ and let $f: M\rightarrow S^p$ be a smooth map. I want to unerstand the proof of the following lemma which occurs in Milnor's 'Topology from the differentiable viewpoint' in the context of Pontryagin-Thom construction.

Lemma 2: If $y$ is a regular value of $f$, and $z$ is sufficiently close to $y$, then $f^{-1}(z)$ ist framed cobordant to $f^{-1}(y)$.

I think I understand most of the proof but I'm stuck with the first sentence.

Proof: Since the set $f(C)$ of critical values is compact, we can choose (...).

Why is that true?

Answer 1

Since both $M$ and $S^p$ are compact, you can choose finite atlases $\{\mathcal{U}_i\}$ and $\{\mathcal{V}_j\}$ for them, where each $\mathcal{U}_i$ is an open subset of $\mathbb{R}^n$ and each $\mathcal{V}_j$ is an open subset of $\mathbb{R}^p$.

Then, if $f_{ij} \colon \mathcal{U}_i \to \mathcal{V}_j$ is the local expression of $f \colon M \to S^p$ in the charts $\mathcal{U}_i$ and $\mathcal{V}_j$, the local expression of the differential map $df \colon TM \to TS^p$ is simply given by the matrix of partial derivatives of $f_{ij}$, and the set of critical points is given by the set of points in $\mathcal{U}_i$ where such a matrix has no maximal rank.

This amount to the vanishing of a finite number of minors, hence the set of critical points of $f_{ij}$ in $\mathcal{U}_i$ is closed. Since there are only finitely many charts, the whole set $C \subset M$ of critical points of $f$ is closed, too. Therefore $C$ is compact, since it is a closed subset of the compact manifold $M$.

Finally, $f \colon M \to S^p$ is continuous, hence the image $f(C)$ of the compact set $C$ is compact and we are done.