Q:

How to show that a regular manifold is itself a manifold?

A set $S$ of a manifold $N$ of dimension $n$ is a regular submanifold of dimension $k$ if for every $p \in S$ there is a coordinate neighborhood $(U,\phi)=(U,x^1,...,x^n)$ of $p$ in the maximal atlas of $N$ such that $U \bigcap S$ is defined by the vanishing of $n-k$ of the coordinate . By renumbering the coordinates , we may assume that these $n-k$ coordinate functions are $x^{k+1},...,x^n$ .  
We call such a chart $(U,\phi)$ in $N$ an adapted chart relative to $S$ . On $U \bigcap S$ , $\phi=(x^1,...,x^k,0,...,0)$ . Let $$\phi_S:U\bigcap S \to R^k$$
  be the restriction of the first $k$ components of $\phi$ to $U \bigcap S$ , that is , $\phi_S=(x^1,...,x^k)$. Note that $(U \bigcap S ,\phi_S)$ is a chart for $S$ in the subspace topology .  

The definition above was in Loring W.Tu's book . But I can not see why $(U \bigcap S,\phi_S)$ is a chart for $S$ . Since both $x^i$ are continuous on $U$ , so it is also continuous on $U \bigcap S$ and it is also one-to-one correspondence. To show $\phi_S$ is a chart , I need to show $\phi_S$ is an open mapping . But I can't even show $\phi_S(U \bigcap S)$ is an open set in $R^k$ . 

A:

First show that $\Phi_S(U \bigcap S) = \phi(U) \bigcap \mathbb R^k$, then use that $\phi(U) \subset \mathbb R^n$ is open, then use the subspace topology on $\mathbb R^k \subset \mathbb R^n$.

