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

Prove the following:

Theorem.
Let $f colon mathbfR^n+k to mathbfR^n$ be of class $C^r$.
Let $M$ be the set of all $mathbfx$ such that $f(mathbfx) = 0$.
Assume that $M$ is non-empty and that $Df(mathbfx)$ has rank $n$ for $mathbfx in M$.
Then $M$ is a $k$-manifold without boundary in $mathbfR^n+k$.
Furthermore, if $N$ is the set of all $mathbfx$ for which
$$
f_1(mathbfx) = dotsb = f_n-1(mathbfx) = 0
quadtextandquad
f_n(mathbfx) geq 0
$$
and if the matrix
$$
partial (f_1, dotsc, f_n-1) / partial mathbfx
$$
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 $alpha_0 : Vsubseteq mathbbR^k to U_(p_0, q_0) subseteq mathbbR^n+k$, where $V$ is open in $mathbbR^k$ and $U_(p_0, q_0)$ is open in $M$, s.t

• $alpha in C^r$




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




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

Proof:

For
$$M = (p,q) in mathbbR^n times mathbbR^k ,$$
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 $F: mathbbR^n+k to mathbbR^n+k$ by
$$F(p,q) = (f(p,q), q).$$

Since
Observe the followings:

• the domain of $F$ is open,


• $Fin C^r$ (since f is a composite function of $f,i in C^r$),


• $det (DF) = det (fracpartial fpartial p )$ - 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 $mathbbR^n+k$ s.t $F|_P_0 times Q_0 : P_0 times Q_0 to F(P_0 times Q_0)subseteq mathbbR^n+k$ 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 $0times T_0$ 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 $mathbbR^n+k$, 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

$alpha_0: G^-1(G(0 times T_0) cap M) subseteq mathbbR^k to G(0 times T_0) cap M$ 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 $mathbbR^k$ ?


3. $alpha_0 in C^r$ since it is the restriction of $G$ to the set $0times T_0$,



4. $Dalpha_0 = DG = (DF)^-1$ 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 $mathbbR^n+k$, 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 $mathbbR^k$ for $alpha_0$ ?

About the point 4:

$alpha_0$ is a function of $qin mathbbR^k$, and it is defined by the function $G$;however, when we find $Dalpha_0$, 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 $Dalpha_0 = D_G$ (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 $mathbbR^n+k$, 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 $mathbbR^n+k$ is the same as the product topology on it, so in a sense, we should be able to express any open set on $mathbbR^n+k$ 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 ?