Why is $(U \cap R, \varphi_R)$ a chart for a regular submanifold $R$?

Question

Let $R$ be a regular submanifold of a manifold $M$ where $\dim(M)=m$ and $\dim(R)=k$. By definition, for every $r \in R$, there is a chart (in $M$) about p, $(U,\varphi) = (U,x^1,...,x^k,...,x^m)$, where $\varphi: U \to \mathbb R^m$, for which $$U \cap R = \{x^{k+1} = ... = x^{m} = 0\}$$ Define $$\varphi_R : U \cap R \to \mathbb R^k, \varphi_R = (x^1,...,x^k)$$

Why is $(U \cap R, \varphi_R)$ a chart for $R$?

We are given:

1. $U$ is open in $M$
2. $\varphi : U \to \varphi(U)$ is a homeomorphism.
3. $\varphi(U)$ is open in $\mathbb R^m$

I think we must show:

1. $U \cap R$ is open in $R$
2. $\varphi_R : U \cap R \to \varphi(U \cap R)$ is a homeomorphism.
3. $\varphi(U \cap R)$ is open in $\mathbb R^k$

I know how to do (1) and (2) unless (3) is assumed in proving either, but I don't know how to do (3).

If (3) is indeed what we must show, then how do we do this?

• All I did so far is prove $\varphi(U \cap R) = \varphi_R(U \cap R) \subseteq \varphi(U) \cap$ '$\mathbb R^k$' (see 4). If the reverse inclusion $\supseteq$ is true, then we are done. Otherwise, I think this comes down to understanding $U \cap R$ (which I don't think is a variety in algebraic geometry), which I apparently do not.

If (3) is not what we must show, then what must we show instead?

$(4)$ By '$\mathbb R^k$', I mean $\mathbb R^k \times \underbrace{\{0\} \times \cdots \times \{0\}}_{\text{m-k times}}$

Thanks in advance!

Answer 1

I think a serious problem with the teaching methods of differential geometry, particularly manifolds, is that sometimes people use sloppy/hand-waving/difficult notations. This is currently my main issue while trying to learn about manifolds. So, even though what they intend to say is not very complicated, the sloppy notations make it really difficult to follow what they intend to say or prove.

I think what the author means to say is that

$$U \cap R = \{x^{k+1}=\cdots=x^m=0\}$$

is just a notation to say that $$\varphi(U \cap R)=\{(x_1,\cdots,x_m)\in\mathbb{R}^m:x^{k+1}=\cdots=x^m=0 \text{ and } \varphi^{-1}(x_1,\cdots,x_m) \in U\}$$

The later one is obviously isomorphic to $\mathbb{R}^k \subset \mathbb{R}^n$ intersected with $\varphi(U)$.

So, $\varphi(U \cap R)=\varphi(U)\cap\mathbb{R}^k$ where $\varphi(U)$ is open in $\mathbb{R}^m$ because $(U,\varphi)$ is a chart in $M$. By the definition of subspace topology, $\varphi(U\cap R)$ is open in $\mathbb{R}^k$.