# Question about a lemma from Milnor's Topology from the Differentiable Viewpoint

I've been reading John W. Milnor's Topology from the Differentiable Viewpoint for some time and currently I'm stuck at a little lemma. I would appreciate if someone can clarify it to me. The details are as follows:

Define $H^m = \{(x_1,\ldots, x_m) \in \mathbb{R}^m | x_m\geq 0\}$. The boundary of this set is denoted as $\partial H^m$. The lemma which I'm finding hard to work through states that if $f:H^m\to \mathbb{R}^n$ ($m>n$) is a smooth map and if $y \in \mathbb{R}^n$ is a regular value for both $f$ and $f|_{\partial H^m}$ then the set of all pre-images of $y$ under $f$ (i.e. $f^{-1}(y)$) is a smooth $(m-n)$-manifold with boundary $\partial (f^{-1}(y)) = f^{-1}(y)\cap \partial X$. [Lemma 4, ch. 2].

To prove this a point $\bar{x}$ was taken from $f^{-1}(y)\cap \partial H^m$ and another smooth map $g:U \to \mathbb{R}^n$ was constructed on a neighborhood $U$ of $\bar{x}$ which agrees with $f$ on $U \cap H^m$. Then it was stated that we can replace $U$ by a sufficiently small neighborhood around $\bar{x}$ on which $g$ will have no critical point. Hence $g^{-1}(y)$ will be a smooth $(m-n)$-manifold. So, my first question is:

Why does there exist a neighborhood of $\bar{x}$ on which $g$ will have no critical point? (And, is this true for any smooth map from a neighborhood of $\bar{x}$ to $\mathbb{R}^n$?)

Next it is noted that the projection map $\pi:g^{-1}(y)\to\mathbb{R}$, $(x_1,\ldots,x_m)\mapsto x_m$ has $0$ as a regular value. Though I can understand why the tangent space $Tg^{-1}(y)_x$ (for some $x \in \pi^{-1}(0)$) is equal to the null space of $dg_x=df_x$ (this follows from a previous lemma), I can't understand why the null space of $df_x$ cannot be contained in $\mathbb{R}^{n-1}\times 0$. And that is my second question:

How does the condition of regularity of $f|_{\partial H^m}$ at $x$ imply that $\ker df_x \nsubseteq \mathbb{R}^{n-1}\times 0$?

For the first question, you should prove the proposition:

If $f: M \to N$ is a smooth map, then the set of regular values of $f$ is an open set.

Sketch of the proof. Let $x$ be a regular point. The derivative $df_x$ is surjective, so the determinant of the Jacobian matrix is nonzero. Every regular value in the set of regular values has this nonzero Jacobian property, so this set is an open set in $N$. $\square$

Now use the continuity of $f$ to look at the preimage of this set. It's a set of only regular values containing $x$. So just intersect $U$ with this preimage if $U$ has critical points, and the resulting intersection will be an open set with only regular points.

We have $$\ker (d(f \vert_{\partial H^m})_x ) = \ker (df_x) \cap \partial H^m.$$ This follows from $TH^m = \mathbf{R}^m$ (being an $m$-dimensional subspace of $\mathbf{R}^m$) and $T(\partial H^m)_x = \mathbf{R}^{m - 1} \times \{ 0 \} = \partial H^m$ (being the image of the local parametrization around $x \in \partial H^m$ given by $(x_1, ..., x_{m - 1}) \mapsto (x_1, ..., x_{m - 1}, 0)$) and the diagram of differentials as usually done in Milnor's book.
So if $\ker (d f_x) \subset \partial H^m$, then $\ker (d(f\vert_{\partial H^m})_x) = \partial H^m$; thus $d(f\vert_{\partial H^m})_x$ would be the zero map and $x$ not a regular value of $f \vert_{\partial H^m}$.