# A $k+1$ Manifold whose boundary is the solution set to the equation $f(\vec{x})=\vec{0}$

Let $f: \mathbb{R}^{n+k} \to \mathbb{R}^n$ be of class $C^r$.

Let $f_1, ..., f_n$ be components of $f$. Define $$M=\{\vec{x} | f(\vec{x})=\vec{0}\},$$ $$N=\{\vec{x} | f_1(\vec{x})=0, ..., f_{n-1}(\vec{x})=0\, f_n(\vec{x})\geq 0\}.$$

Assume that

(1). the Jacobian matrix $Df(\vec{x})$ has rank $n$ for each $\vec{x}\in M$ and that

(2). the matrix $$\frac{\partial (f_1, ..., f_{n-1})}{\partial (\vec{x})}$$ has rank $n-1$ for each $\vec{x}\in N$.

Prove that $N$ is a $k+1$-manifold whose boundary is $M$.

Attempt I've proved that $M$ is a $k$-manifold.

Define $F: \mathbb{R}^{n+k} \to \mathbb{R}^{n+k}$ by

$$F(x_1, ..., x_k, x_{k+1}, ..., x_{n+k})=(x_1, ..., x_k, f_1(\vec{x}), ..., f_n(\vec{x})).$$

The given $\vec{x}\in M$, we know that $DF(\vec{x})$ has rank $n+k$, hence is non-singular, since $Df(\vec{x})$ has rank $n$. So $F$ is a diffeomorphism of class $C^r$ (Inverse Function Theorem) of a neighborhood $A$ of $\vec{x}$ in $\mathbb{R}^{n+k}$ with an open set $B$ in $\mathbb{R}^{n+k}$.

$F$ carries the open set $V=A\cap M$ of $M$ onto the open set $B\cap (\mathbb{R}^k \times {0}^n)$ of $\mathbb{R}^k \times {0}^n$.

Let $\pi: \mathbb{R}^k \times {0}^n \to \mathbb{R}^k$ be the projection from $\mathbb{R}^k \times {0}^n$ onto $\mathbb{R}^k$. Then $\pi$ is an open map, hence carries $B\cap (\mathbb{R}^k \times {0}^n)$ onto an open set $U$ in $\mathbb{R}^{k}$.

Now it is easy to verify that $\alpha: \mathbb{R}^k \to \mathbb{R}^{n+k}$ defined by $\alpha=(\pi \circ F)^{-1}=F^{-1}\circ \pi^{-1}$ is a bijection of class $C^r$ whose inverse is continuous on $V$. Moreover, $D\alpha (\vec{x})$ has rank $k$ for $\vec{x}\in M$. Therefore $\alpha$ is a coordinate patch, so $M$ is a $k$-manifold.

However, after manipulation with the definition of $N$, I cannot find a coordinate patch $g: \mathbb{R}^{k+1} \to \mathbb{R}^{n+k}$ on $N$.

Can anyone suggest how to prove that $N$ is a $k+1$-manifold? Thank you!