I have some questions about the arguments used in following excerpt in J. Milnor's " Topology from the differentiable view point":
We have following setting: $f: H^m \to N^n$ is a smooth map from a $m$-manifold with boundary $\partial H^m$ to a $n$-manifold without boundary. Set $y \in N_n $ regular for $f$ and the restriction $f \vert _{\partial N}$. This means that $dim T_x f= dim_x (f \vert _{\partial H^m}) =n$ for every $x \in f^{-1}(y)$.
Since the problem is local we assume $N^n= \mathbb{R}^n$.
My questions are:
Let $x \in \pi^{-1}(0) \subset f^{-1}(y)$.
1.: Why the tangent space $T_x \pi$ equals to the null space (= kernel) of the tangential map $dg_x = df: \mathbb{R}^m \to \mathbb{R}^n$? (remark: $dg_x = T_xg: T_xH^m \to T_{g(x)}\mathbb{R}^n$ where $T_xH^m= \mathbb{R}^m$ since $H^m$ is a $m$-manifold and $T_{g(x)}\mathbb{R}^n = \mathbb{R}^n$
- Why the fact that $f \vert _{\partial H^m}$ and $f$ are regular implies that $ker(T_x g) \not \subset \mathbb{R}^{m-1} \times \{0\}$?
We know that since $f$ and $f \vert _{\partial H^m}$ regular, then $ker(T_x f)= m-n$ and $ker(T_x (f \vert _{\partial H^m}))= m-1-n$ by linear algbra but I don't see how it should imply $\pi(ker(T_x f)) \neq 0$.
This should answer question 1. I am not sure, immediately, how to answer your second question. I'll make an edit if I think of something.