Completing the proof of that the set of points where $f(x)=0 $ is a $k$-manifold

Question

In the book of Analysis on Manifolds by Munkres, at page 208 question 2, it is asked that

Prove the following:

Theorem. Let $f \colon \mathbf{R}^{n+k} \to \mathbf{R}^n$ be of class $C^r$. Let $M$ be the set of all $\mathbf{x}$ such that $f(\mathbf{x}) = 0$. Assume that $M$ is non-empty and that $Df(\mathbf{x})$ has rank $n$ for $\mathbf{x} \in M$. Then $M$ is a $k$-manifold without boundary in $\mathbf{R}^{n+k}$. Furthermore, if $N$ is the set of all $\mathbf{x}$ for which $$ f_1(\mathbf{x}) = \dotsb = f_{n-1}(\mathbf{x}) = 0 \quad\text{and}\quad f_n(\mathbf{x}) \geq 0 $$ and if the matrix $$ \partial (f_1, \dotsc, f_{n-1}) / \partial \mathbf{x} $$ has rank $n-1$ at each point of $N$, then $N$ is a $k+1$ manifold, and $\partial N = M$.

(Hint: Examine the proof of the implicit function theorem)

(Original images here and here.)

I have given a partial answer to the first part of this question only, where there are some part that I couldn't figure out how to get around, and some points where I do not know whether it is acceptable. The problematics part being with the title "The problems begin" toward the end of the answer.

Partial Solution:

To show: $M$ is a k-manifold without boundary, so given $(p_0, q_0) \in M$, we find a map $\alpha_0 : V\subseteq \mathbb{R}^k \to U_{(p_0, q_0)} \subseteq \mathbb{R}^{n+k}$, where $V$ is open in $\mathbb{R}^k$ and $U_{(p_0, q_0)}$ is open in $M$, s.t

• $\alpha \in C^r$

• $\alpha^{-1}$ exists, and of class $C^r$

• $D\alpha (p,q)$ has rank at least $k$ for all $(p,q) \in V$.

Proof:

For $$M = \{(p,q) \in \mathbb{R}^n \times \mathbb{R}^k | f(p,q) = 0\},$$ let $(p_0, q_0) \in M$ be arbitrary but fix (since $M$ is non-empty, we can do this).

First, we are going to construct an open neighbourhood of $(p_0, q_0)$.

Let define $F: \mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ by $$F(p,q) = (f(p,q), q).$$

Since Observe the followings:

• the domain of $F$ is open,
• $F\in C^r$ (since f is a composite function of $f,i \in C^r$),
• $\det (DF) = \det (\frac{\partial f}{\partial p} )$ - in particular, $\det (DF (p_0, q_0)) \not = 0$ by our hypothesis. (Of course, in here we are assuming that Df has rank n for the the first n argument, wlog.)

Then, by the inverse function theorem, there exists an open neighbourhood $P_0 \times Q_0$ of $(p_0,q_0)$ in $\mathbb{R}^{n+k}$ s.t $F|_{P_0 \times Q_0} : P_0 \times Q_0 \to F(P_0 \times Q_0)\subseteq \mathbb{R}^{n+k}$ is one-to-one and onto.Moreover, $$G: F(P_0 \times Q_0) \to P_0 \times Q_0 \in C^r.$$

Lets make 2 observations:

$$F(p_0, q_0) = (0, q_0),$$ and $$F(f(p_0,q_0), q_0) = G(0,q_0) = (p_0, q_0).$$

Since $F(P_0 \times Q_0) \ni (0,q_0)$ open, there exists an open neighbourhood $0\times T_0$ of $(0,q_0)$ that is contained in $F(P_0 \times Q_0)$.

Since $F \in C^r$, $G(0 \times T_0)$ is open in $\mathbb{R}^{n+k}$, hence $G(0 \times T_0) \cap M$ is open in $M$ (by the subspace topology).

The problems begin:

If we define our coordinate patch $\alpha_0$ as

$\alpha_0: G^{-1}(G(0 \times T_0) \cap M) \subseteq \mathbb{R}^k \to G(0 \times T_0) \cap M$ by $$\alpha (q) = G(0,q).$$

we can see that

1. $\alpha_0 (q_0) = G(0,q_0) = (p_0, q_0),$
2. Does $G^{-1}(G(0 \times T_0) \cap M)$ open in $\mathbb{R}^k$ ?
3. $\alpha_0 \in C^r$ since it is the restriction of $G$ to the set $0\times T_0$,

4. $D\alpha_0 = DG = (DF)^{-1}$ has rank $n+k$,

5. $\alpha_0^{-1}$ is continuous since $F \in C^r$.

About the point 2:

Since $M$ has measure zero in $\mathbb{R}^{n+k}$, it is not open and hence $G(0 \times T_0) \cap M$ is not open either, so the inverse image cannot open too, but if so, how are we going to find an open set in $\mathbb{R}^k$ for $\alpha_0$ ?

About the point 4:

$\alpha_0$ is a function of $q\in \mathbb{R}^k$, and it is defined by the function $G$;however, when we find $D\alpha_0$, we need to take the partial derivatives of $G$ after plugging $0$ into the first $n$ arguments, so I'm not sure how can we make sure ourselves that $D\alpha_0 = D_G$ (since we are first plugging the numbers and then taking the partial derivatives, and not the reverse).

Flaws:

1. The inverse function theorem gives as an open neighbourhood of $(p_0,q_0)$ in $\mathbb{R}^{n+k}$, but this set does not have to be in such a simple form that as $P_0 \times Q_0$, which is what Munkres claim in the proof of the implicit function theorem. Even though, the standard topology on $\mathbb{R}^{n+k}$ is the same as the product topology on it, so in a sense, we should be able to express any open set on $\mathbb{R}^{n+k}$ by the union of such simple open sets, I'm not sure whether does this cause any problem.

Question: (in summary)

First of all, how to get around with the problems that I have pointed out in my partial answer.

Secondly, as I have mentioned explicitly, there is a flaw both in my and Munkres' proof (as far as I can tell), but is it really a flaw, or is there as way to get around it ?

Thirdly, it the rest of my partial answer correct / have any flaw in it which I couldn't see ?