MHB Munkres' 'Analysis on Manifolds' Question.

[caffeinemachine]

In Munkres' 'Analysis on Manifolds' on pg. 208 there's a question which reads:QUESTION: Let $f:\mathbb R^{n+k}\to \mathbb R^n$ be of class $\mathscr C^r$.
Let $M$ be the set of all the points $\mathbf x$ such that $f(\mathbf x)=\mathbf 0$ and $N$ be the set of all the points $\mathbf x$ such that $$f_1(\mathbf x)=\cdots=f_{n-1}(\mathbf x)=0\text{ and } f_n(\mathbf x)\geq 0$$
Assume $M$ is non-empty.1) Assume $\text{rank} Df(\mathbf x)=n$ for all $\mathbf x\in M$ and show that $M$ is a $k$-manifold without boundary in $\mathbb R^{n+k}$.2) Assume that the matrix $\displaystyle\frac{\partial(f_1,\ldots,f_{n-1})}{\partial \mathbf x}$ has rank $n-1$ for all $\mathbf x\in N$ and show that $N$ is a $(k+1)$-manifold with boundary in $\mathbb R^{n+k}$.
___I am trying to show $(2)$ and I am not sure if the hypothesis of $(1)$ is required to do that.I have approached this question using the constant rank theorem which dictates:

Constant Rank Theorem: Let $U$ be open in $\mathbb R^n$ and $\mathbf a$ be any point in $U$. Let $f:U\to \mathbb R^m$ be a function of class $\mathscr C^p$ such that $\text{rank } Df(\mathbf z) =r$ for all $\mathbf z\in U$. Then there exist open sets $U_1,U_2\subseteq U$ and $V\subseteq \mathbb R^m$ such that $\mathbf a\in U_1$ and $f(\mathbf a)\in V$, and $\mathscr C^p$-diffeomorphisms $\phi:U_1\to U_2$ and $\psi:V\to V$ such that $$(\psi\circ f\circ \phi^{-1})(\mathbf z)=(z_1,\ldots,z_r,0,\ldots,0)$$
for all $\mathbf z\in U_2$.
___
What I did was define a function $g:\mathbb R^{n+k}\to \mathbb R^{n-1}$ as $$g(\mathbf x)=(f_1(\mathbf x),\ldots,f_{n-1}(\mathbf x))$$
Then $\text{rank }Dg(\mathbf x)=n-1$ for all $\mathbf x\in N$.
Let $\mathbf z_0\in N$.
I can show that there exists an open set $U\subseteq \mathbb R^{n+k}$ such that $\mathbf z_0\in U$ and $\text{rank }Dg(\mathbf z)=n-1$ for all $\mathbf z\in U$.
Thereby, using the conastant rank theorem I get $U_1, U_2,\psi$ and $\phi$ such that $(\psi\circ g\circ\phi^{-1})(\mathbf x)=(x_1,\ldots,x_{n-1})$
Can somebody guide me what to do from here?

 

[caffeinemachine]

My approach to solve $(2)$ shall be clear by my solution of $(1)$:

Let $\mathbf a\in M$.
We know that there exists $U$ open in $\mathbb R^{n+k}$ such that $\mathbf a\in U$ and $\text{rank }Df(\mathbf x)=n$ for all $\mathbf x\in U$.
By the Constant Rank Theorem(see OP) there exists open sets $U_1$ and $U_2$ in $\mathbb R^{n+k}$ and $V$ in $\mathbb R^n$ such that $\mathbf a\in U_1\subseteq U_1$ and $f(\mathbf a)=\mathbf \in V$, along with diffeomorphisms $\phi:U_1\to U_2$ and $\psi:V\to V$ satisfying
$$(\psi\circ f\circ \phi^{-1})(\mathbf x) =(x_1,\ldots,x_n)$$
for all $\mathbf x\in U_2$.
Say $\psi(\mathbf 0)=(t_1,\ldots,t_n)$ and define $S=\{(t_1,\ldots,t_n,z_1,\ldots,z_k):z_i\in \mathbb R\}\cap U_2$.

Claim 1: $\phi^{-1}(S)=M\cap U_1$.
Proof: Let $\mathbf q=(t_1,\ldots,t_n,z_1,\ldots,z_k)$ be in $S$.
Then $\phi^{-1}(\mathbf q)$ obviously lies in $U_1$.
We now show that $\phi^{-1}(\mathbf q)$ lies in $M$.
Note that $(\psi\circ f\circ \phi^{-1})(\mathbf q)=(t_1,\ldots,t_n)$.
This gives $(f\circ \phi^{-1})(\mathbf q)=\psi^{-1}(t_1,\ldots,t_n)=\mathbf 0$.
This means that $f(\phi^{-1}(\mathbf q))=\mathbf 0$ and hence $\phi^{-1}(\mathbf q)$ is in $M$.
For the reverse containment assume that $\mathbf q\in M\cap U_1$.
Then $\mathbf q=\phi^{-1}(\mathbf s)$ for some $\mathbf s\in U_2$.
Also, $f(\mathbf q)=0$ since $\mathbf q\in M$.
Thus $(f\circ\phi^{-1})(\mathbf s)=\mathbf 0$.
Operating $\psi$ on both the sides we get $(\psi\circ f\circ \phi^{-1})(\mathbf s)=\psi(\mathbf 0)$.
But the LHS of the last equation is $(s_1,\ldots,s_n)$ and the RHS is $(t_1,\ldots,t_n)$.
Thus $s_i=t_i$ for $1\leq i\leq n$.
Therefore $\mathbf s\in S$ and $\mathbf q\in\phi^{-1}(S)$.
This settles the claim.

Now define $T=\{(z_1,\ldots,z_k)\in\mathbb R^k: (t_1,\ldots,t_n,z_1,\ldots,z_k)\in S\}$.

Claim 2: $T$ is open in $\mathbb R^k$.
Proof: Define $g:\mathbb R^k\to \mathbb R^{n+k}$ as $$g(z_1,\ldots, z_k)=(t_1,\ldots,t_n,z_1,\ldots,z_k)$$
Clearly $g$ is injective and continuous.
We now show that $g^{-1}(U_2)=T$.
Note that $g^{-1}(U_2)=g^{-1}(S)$.
Let $\mathbf q\in S$.
Say $\mathbf q=(t_1,\ldots,t_n,q_1,\ldots,q_k)$ and it is obvious that $g^{-1}(\mathbf q)\in T$.
Now let $g^{-1}(\mathbf q)\in T$ for some $q\in \mathbb R^{n+k}$.
We need to show that $\mathbf q\in U_2$.
Say $g^{-1}(\mathbf q)=(b_1,\ldots,b_k)$.
Then $\mathbf q=(t_1,\ldots,t_n,b_1,\ldots,b_k)\in S$ and thus $\mathbf q\in U_2$.

So we have shown that $T=g^{-1}(U_2)$.
Now since $g$ is a continuous function and $U_2$ is open in $\mathbb R^{n+k}$, we infer that $T$ is open in $\mathbb R^k$ and the claim is settled.
Now define a function $\alpha:T\to M\cap U_1$ as $$\alpha(\mathbf z)=\phi^{-1}\circ g(\mathbf z)$$
It is a trivial matter to verify that $\alpha$ is a coordinate patch about the point $\mathbf a$ in $M$ and the proof is complete.

 

[ThePerfectHacker]

In Munkres' 'Analysis on Manifolds' on pg. 208 there's a question which reads:QUESTION: Let $f:\mathbb R^{n+k}\to \mathbb R^n$ be of class $\mathscr C^r$.
Let $M$ be the set of all the points $\mathbf x$ such that $f(\mathbf x)=\mathbf 0$ and $N$ be the set of all the points $\mathbf x$ such that $$f_1(\mathbf x)=\cdots=f_{n-1}(\mathbf x)=0\text{ and } f_n(\mathbf x)\geq 0$$
Assume $M$ is non-empty.1) Assume $\text{rank} Df(\mathbf x)=n$ for all $\mathbf x\in M$ and show that $M$ is a $k$-manifold without boundary in $\mathbb R^{n+k}$.do from here?

Here is a way to do it, perhaps it be useful to you.

Let $x\in M$. We will produce an open set $U\subset \mathbb{R}^{n+k}$ containing $x$ and an open set $V \subset \mathbb{R}^k$ together with a diffeomorphism $\varphi: (U\cap M) \to V$. This will prove, by definition, that $M$ is a $k$-manifold.

Let $x\in M$. By assumption the linear map $Df(x):\mathbb{R}^{n+k} \to \mathbb{R}^n$ has full rank. It follows from here by the rank-nullity theorem that $\ker Df(x)$ is a $k$-dimensional subspace of $\mathbb{R}^{n+k}$. To ease notation let $N = \ker Df(x)$, called the nullspace, of $Df(x)$.

Choose any basis for $N$, say $\{a_1,a_2,...,a_k\}$. This basis for $N$ can be extended to a basis of $\mathbb{R}^{n+k}$ as $\{a_1,a_2,...,a_k,b_1,b_2,...,b_n\}$. Use this basis for $\mathbb{R}^{n+k}$ to define the linear map $L:\mathbb{R}^{n+k} \to \mathbb{R}^k$ by $L(a_j) = e_j$ and $L(b_j) = 0\in \mathbb{R}^k$ where $e_j = (0,0,...,1,...0)\in \mathbb{R}^k$. If we restrict this linear map to $N$ we get a linear map $L|_N : N \to \mathbb{R}^k$ that is invertible. The reason for this is simply by choosing bases and writing out the corresponding matrix. If we use the basis $(a_j)$ for $N$ and $(e_j)$ for $\mathbb{R}^k$ then the corresponding matrix is the $(k\times k)$ identity matrix which is clearly invertible. To summarize $L:\mathbb{R}^{n+k}\to \mathbb{R}^k$ is a linear map that is invertible when restricted to $N$.

Define the following function, $g:\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ by,
$$ g(p) = \bigg(f(p),L(p)\bigg)$$
Just to make sure we understand this function let us explain what it is doing. Given a point in $p \in \mathbb{R}^{n+k}$ if we apply $f$ the result $f(p)$ is an element of $\mathbb{R}^n$. If we apply $L(p)$ the result is an element of $\mathbb{R}^k$. So if we juxtapose $f(p)$ next to $L(p)$ we get an $(n+k)$-tuple of real numbers. The definition above for $g(p)$ is a little abused, rather we have, the codomain $\mathbb{R}^n\times \mathbb{R}^k$ but we write it as $\mathbb{R}^{n+k}$ because we are secretly juxtaposing the coordinates together to get a $(n+k)$-tuple.

This function $g:\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ is clearly a smooth map. This has to do with the fact that $g$ is essentially made out of two the smooth map $Df(x)$, and the linear map $L$, both of which are smooth maps. We next determine the linear map $Dg(x):\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$. Since the (Frechet') derivative of a linear map is itself and $g$ is made out of $f$ and $L$ we have that,
$$ Dg(x)(p) = (Df(x)(p),L(p)) $$
Where we again understand that the meaning of the RHS is that we juxtapose coordinates together to form an $(n+k)$-tuple. We claim that $Dg(x)$ is an invertible linear map. To see this suppose that $p\in \mathbb{R}^{n+k}$ was such a point so that $Dg(p) = 0$, but then it would mean that $Df(x)(p)$ and $L(p) = 0$. Thus, $p\in N$, the nullspace of $Df(x)$, and at the same time $L(p) = 0$, but recall that $L$ when restricted to $N$ was invertible, so that $p = 0$. This shows that only $0$ is in the nullspace of $Dg(x)$ and hence $Dg(x)$ is an invertible linear map.

It now follows from the inverse-function theorem that there exists a neighborhood $U\subset \mathbb{R}^{n+k}$ of $x$ and a neighborhood $\mathcal{O} \subset \mathbb{R}^{n+k}$ of $(0,L(x))$ such that $g: U \to \mathcal{O}$ is a diffeomorphism.

Now $M$ is send onto $\{ 0 \} \times \mathbb{R}^k$ by the map $g$. Thus, we have $g: (U\cap M)\to (\{ 0 \}\times \mathbb{R}^k)\cap \mathcal{O}$.

At this point your intuition should tell you that $(\{ 0 \}\times \mathbb{R}^k)\cap \mathcal{O}$ is an open subset of $\mathbb{R}^k$. To be more precise we let $V$ be the set after killing off the first $n$ entries of $\mathcal{O}$, this is an open subset of $\mathbb{R}^k$ and so we get a diffeomorphism $\varphi: (U\cap M) \to V$ where $V$ is an open subset of $\mathbb{R}^k$.

 

[caffeinemachine]

Here is a way to do it, perhaps it be useful to you.

Let $x\in M$. We will produce an open set $U\subset \mathbb{R}^{n+k}$ containing $x$ and an open set $V \subset \mathbb{R}^k$ together with a diffeomorphism $\varphi: (U\cap M) \to V$. This will prove, by definition, that $M$ is a $k$-manifold.

Let $x\in M$. By assumption the linear map $Df(x):\mathbb{R}^{n+k} \to \mathbb{R}^n$ has full rank. It follows from here by the rank-nullity theorem that $\ker Df(x)$ is a $k$-dimensional subspace of $\mathbb{R}^{n+k}$. To ease notation let $N = \ker Df(x)$, called the nullspace, of $Df(x)$.

Choose any basis for $N$, say $\{a_1,a_2,...,a_k\}$. This basis for $N$ can be extended to a basis of $\mathbb{R}^{n+k}$ as $\{a_1,a_2,...,a_k,b_1,b_2,...,b_n\}$. Use this basis for $\mathbb{R}^{n+k}$ to define the linear map $L:\mathbb{R}^{n+k} \to \mathbb{R}^k$ by $L(a_j) = e_j$ and $L(b_j) = 0\in \mathbb{R}^k$ where $e_j = (0,0,...,1,...0)\in \mathbb{R}^k$. If we restrict this linear map to $N$ we get a linear map $L|_N : N \to \mathbb{R}^k$ that is invertible. The reason for this is simply by choosing bases and writing out the corresponding matrix. If we use the basis $(a_j)$ for $N$ and $(e_j)$ for $\mathbb{R}^k$ then the corresponding matrix is the $(k\times k)$ identity matrix which is clearly invertible. To summarize $L:\mathbb{R}^{n+k}\to \mathbb{R}^k$ is a linear map that is invertible when restricted to $N$.

Define the following function, $g:\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ by,
$$ g(p) = \bigg(Df(x)(p),L(p)\bigg)$$
Just to make sure we understand this function let us explain what it is doing. Given a point in $p \in \mathbb{R}^{n+k}$ if we apply $Df(x)$ to be the result $Df(x)(p)$ is an element of $\mathbb{R}^n$. If we apply $L(p)$ the result is an element of $\mathbb{R}^k$. So if we juxtapose $Df(x)(p)$ next to $L(p)$ we get an $(n+k)$-tuple of real numbers. The definition above for $g(p)$ is a little abused, rather we have, the codomain $\mathbb{R}^n\times \mathbb{R}^k$ but we write it as $\mathbb{R}^{n+k}$ because we are secretly juxtaposing the coordinates together to get a $(n+k)$-tuple.

This function $g:\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ is clearly a smooth map. This has to do with the fact that $g$ is essentially made out of two linear maps, $Df(x)$, and $L$, both of which are smooth maps. We next determine the linear map $Dg(x):\mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$. Since the (Frechet') derivative of a linear map is itself and $g$ is made out of two linear maps we have that,
$$ Dg(x)(p) = (Df(x)(p),L(p)) $$
Where we again understand that the meaning of the RHS is that we juxtapose coordinates together to form an $(n+k)$-tuple. We claim that $Dg(x)$ is an invertible linear map. To see this suppose that $p\in \mathbb{R}^{n+k}$ was such a point so that $Dg(p) = 0$, but then it would mean that $Df(x)(p)$ and $L(p) = 0$. Thus, $p\in N$, the nullspace of $Df(x)$, and at the same time $L(p) = 0$, but recall that $L$ when restricted to $N$ was invertible, so that $p = 0$. This shows that only $0$ is in the nullspace of $Dg(x)$ and hence $Dg(x)$ is an invertible linear map.

It now follows from the inverse-function theorem that there exists a neighborhood $U\subset \mathbb{R}^{n+k}$ of $x$ and a neighborhood $\mathcal{O} \subset \mathbb{R}^{n+k}$ of $(0,L(x))$ such that $g: U \to \mathcal{O}$ is a diffeomorphism.

Now $M$ is send onto $\{ 0 \} \times \mathbb{R}^k$ by the map $g$. Thus, we have $g: (U\cap M)\to (\{ 0 \}\times \mathbb{R}^k)\cap \mathcal{O}$.

At this point your intuition should tell you that $(\{ 0 \}\times \mathbb{R}^k)\cap \mathcal{O}$ is an open subset of $\mathbb{R}^k$. To be more precise we let $V$ be the set after killing off the first $n$ entries of $\mathcal{O}$, this is an open subset of $\mathbb{R}^k$ and so we get a diffeomorphism $\varphi: (U\cap M) \to V$ where $V$ is an open subset of $\mathbb{R}^k$.

Thanks for participating.
I haven't studied your proof in detail but it seems you have proved (1) and not (2).
What I was looking for was a solution to (2).
I had already provided a proof for (1) in my second post.

 

[ThePerfectHacker]

Here is a nice generalization of your theorem that you might like.
This is a standard theorem in differencial topology.

Let $M\subset \mathbb{R}^r$ be an $m$-dimensional manifold and $N\subset \mathbb{R}^{\ell}$ be an $n$-dimensional manifold with $m > n$. Let $f:M\to N$ be a smooth map. We say a point $p\in M$ is a regular point if the linear map $Df(x):T_pM\to T_{f(p)}N$ is full-rank i.e. is onto. A point $q\in N$ is a regular value if every point of $f^{-1}(\{q\})$ is a regular point.

Theorem: If $f:M\to N$ is smooth and $q\in N$ is regular value then $f^{-1}(\{q\})$ is $(m-n)$ dimensional manifold.

Note how this result generalizes your theorem.