Sard's theorem and the null space of the higher dimensional manifold.

Question

I was reading Milnor's Topology from a Differentiable Viewpoint, where he talks about Sard's theorem (chapter 2). Milnor proposes a lemma:

Lemma 1. If $f: M \rightarrow N$ is a smooth map between manifolds of dimension $m \geq n$, and if $ y \in N$ is a regular value, then the set $f^{-1}(y) \subset M$ is a smooth manifold of dimension $m - n$.

I was not clear how he proves that the set $f^{-1}(y)$ is of dimension $m - n$?

Milnor suggests that part of the proof lies in the fact that the null space $\mathscr{R} \subset TM_x$ of $df_x$ will be an $m - n$ dimensional vector space, but I was not clear how this proves this point.

I tried to play around with this--probably way off. So if I have two manifolds that are spheres, one in 5 dimension and the second in 3 dimensions. So $M: S^5, N: S^3$. Now there is a mapping $f: M \rightarrow N$. According to the point above, the null space of the tangent space of $M$ should be 2 dimensional.

If we set $$ f = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\\ \end{bmatrix} $$

Then the derivative of $f$ is a similar looking matrix

$$ df = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\\ \end{bmatrix} $$

This is the a three dimensional identity matrix, so why is the null space of this derivative equal to 2? That does not seem to make sense.

Any help would be appreciated.

Answer 1

I don't really get which mapping $f\colon\mathbb{S}^5\rightarrow\mathbb{S}^2$ you are dealing with.

First of all, in any cases $\mathrm{d}_xf$ is not a square matrix, it is a $5\times 2$ matrix, precisely because $\mathbb{T}_x\mathbb{S}^5$ has dimension $5$ and $\mathbb{T}_{f(x)}\mathbb{S}^2$ has dimension $2$.

Then, if you are considering something like $f(x,y,z,0,0,0)=(x,y,z)$ or $f(x,y,z,t,u,v)=(x,y,z)$, then none of them is a mapping from $\mathbb{S}^5$ into $\mathbb{S}^2$. The first one is not define on the whole $\mathbb{S}^5$ and the codomain of the last one is not $\mathbb{S}^2$. For example, $(0,0,0,0,1)\in\mathbb{S}^5$ but $(0,0,0)\not\in\mathbb{S}^2$.

Regarding Milnor's claim : By definition, $y\in N$ is a regular value of $f\colon M\rightarrow N$ if and only if for each $x\in f^{-1}(\{y\})$, $f$ is a submersion at $x$. Namely, $y\in N$ is a regular value of $f\colon M\rightarrow N$ if and only if for each $x\in f^{-1}(\{y\})$, $\mathrm{d}_xf\colon T_xM\rightarrow T_{f(x)}N$ is a surjective linear map. Notice, that $T_xM$ has dimension $m$ and $T_xN$ has dimension $n$, then using rank-nullity theorem: $$\dim(\ker(\mathrm{d}_xf))=m-n.$$ To conclude, notice that $\ker(\mathrm{d}_xf)$ is precisely $T_xf^{-1}(\{y\})$.