Smooth Manifold Chart Lemma for Manifolds with Boundary

Question

Let's start with Lemma 1.35 (Smooth Manifold Chart Lemma for Manifolds Without Boundary) in John Lee's Textbook "Introduction to Smooth Manifolds" (Second Edition). The precise statement is:

Let $M$ be a set and $\{U_\alpha\}_{\alpha\in J}$ be a collection of subsets of $M$, along with maps $\varphi_\alpha:U_\alpha\to\mathbb R^n$, such that the following properties are satisfied:

(i) $\forall \alpha\in J$: $\varphi_\alpha$ is an injective map and $\varphi_\alpha(U_\alpha)$ is open in $\mathbb R^n$.

(ii) $\forall \alpha,\beta\in J$: the sets $\varphi_\alpha(U_\alpha\cap U_\beta)$ and $\varphi_\beta(U_\alpha\cap U_\beta)$ are open in $\mathbb R^n$.

(iii) $\forall\alpha,\beta\in J$: $U_\alpha\cap U_\beta\neq \emptyset \quad \Rightarrow \quad \varphi_\beta\circ\varphi_\alpha^{-1}:\varphi_\alpha(U_\alpha\cap U_\beta)\to \varphi_\beta(U_\alpha\cap U_\beta)$ is smooth.

(iv) Countably many of the sets $U_\alpha$ cover $M$.

(v) $ \left. \begin{array}{c} p,q\in M\\ p\neq q \end{array} \right\} \quad \Rightarrow \quad \left\{ \begin{array}{c} \exists \alpha\in J\text{ such that } p,q\in U_\alpha,\quad\text{ or}\\ \exists \alpha,\beta\in J\text{ such that } p\in U_\alpha, q\in U_\beta \text{ and } U_\alpha\cap U_\beta=\emptyset \end{array} \right. $

Then $M$ has a unique manifold structure such that each pair $(U_\alpha,\varphi_\alpha)$ is a smooth chart.

Let $\mathcal B=\{\varphi_\alpha^{-1}(V):\alpha\in J, V\text{ open in } \mathbb R^n\}$.

From $(iv)$ we see that the elements of $\mathcal B$ cover $M$. Now let $\varphi_\alpha^{-1}(V)$ and $\varphi_\beta^{-1}(W)$ be two elements of $\mathcal B$, where $V$ and $W$ are open in $\mathbb R^n$. To show that $\mathcal B$ forms a basis, it is enough to show that $ \varphi_\alpha^{-1}(V)\cap\varphi_\beta^{-1}(W)$ itself lies in $\mathcal B$. Note that \begin{equation*} \varphi_\alpha^{-1}(V)\cap \varphi_\beta^{-1}(W)=\varphi_\alpha^{-1}\Big(V\cap(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)\Big) \tag{1} \end{equation*} But by (iii), $\varphi_\beta\circ\varphi_\alpha^{-1}$ is continuous, and therefore $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\varphi_\alpha(U_\alpha\cap U_\beta)$. By (ii), $\varphi_\alpha(U_\alpha\cap U_\beta)$ is open in $\mathbb R^n$ and therefore $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\mathbb R^n$. Using this in $(1)$, we immediately see that $\varphi_\alpha^{-1}(V)\cap\varphi_\beta^{-1}(W)$ is in $\mathcal B$. This settles the claim.

The maps $\varphi_\alpha:U_\alpha \to \mathbb{R}^n$ are automatically continuous.

To see they are homeomorphisms with the images, it is equivalent to show that $\varphi_\alpha$ is an open map. To this purpose, it is sufficient to show that $\varphi_\alpha(B)$ is open in $\mathbb{R}^n$ whenever $B$ is an element of $\mathcal{B}$ contained in $U_\alpha$. An arbitrary element of $\mathcal {B}$ is of the form $\varphi^{-1}_\beta(W)$ with $W$ open in $\mathbb{R}^n$. We have $\varphi_\alpha(\varphi^{-1}_\beta(W))=\varphi_\alpha\circ\varphi_\beta^{-1}(W)$ is open in $\varphi_\alpha(U_\alpha \cap U_\beta)$ and thus in $\mathbb{R}^n$.

Question 1) Lee says that each map $\varphi_\alpha$ is an homeomorphism onto its image "essentially by definition", but according to my argument above we are using again hypothesis (iii) and (ii). So, I would say that the continuity of the $\varphi_\alpha$'s is "essentially by definition" (since we are putting in $\mathcal{B}$ all the counter images of the open subsets of $\mathbb{R}^n$) but not the openness. So, does my argument above (to show that the $\varphi_\alpha$'s are homeomporphism onto their images) use unnecessaryly the hypothesis (iii) and (ii)? In other words, is there a simpler way (that justifies the sentence "essentially by definition") to see that the $\varphi_\alpha$'s are homeomorphism onto their images?

Question n° 2 On page 28 Exercise 1.42 says: Show that Lemma 1.35 holds with $\mathbb{R}^n$ replaced by $\mathbb{R}^n$ or $\mathbb{H}^n$ and "smooth manifold" replaced by "smooth manifold with boundary". I think I can copy the same proof of Lemma 1.35, but when I arrive to the point of showing that $ \varphi_\alpha^{-1}(V)\cap\varphi_\beta^{-1}(W)$ itself lies in $\mathcal B$ I'm in trouble because I cannot show that $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\mathbb{R}^n$. What I know is that $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\varphi_\alpha(U_\alpha\cap U_\beta)$, and this last one can be open in $\mathbb{R}^n$ or $\mathbb{H}^n$. In the latter case I have $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)=\mathbb{H}^n \cap S$ with $S$ open subet of $\mathbb{R}^n$, but the set $\varphi_\alpha^{-1}(S)$ can be greater than the set $\varphi_\alpha^{-1}((\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W))$.

My current thought is that: if in the statement of Lemma 1.35 I change (ii) with

(j) $\forall \alpha,\beta\in J$: the set $\varphi_\alpha(U_\alpha\cap U_\beta)$ is open in $\varphi_\alpha(U_\alpha)$

nothing change in Lemma 1.35 but as regard Exercise 1.42, I have that $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\varphi_\alpha(U_\alpha\cap U_\beta)$, which is open in $\phi_\alpha(U_\alpha)$ and this last one can be open in $\mathbb{R}^n$ or $\mathbb{H}^n$. In the latter case I have $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)=\mathbb{H}^n \cap S$ with $S$ open subet of $\mathbb{R}^n$, but since the image of $\varphi_\alpha$ lies in $\mathbb{H}^n$ I have also $\varphi_\alpha^{-1}(S)=\varphi_\alpha^{-1}((\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W))$.

Is my modification correct? Is my modification necessary? I suspect there are many simple things I'm missing, my apologies for this.

Answer 1

Q1:

It is a philosophical question what Lee means when he says "essentially by definition".

It seems you interpret it in the sense that once it has been shown that $\mathcal B$ is closed with respect to intersecting two members (i.e. that it generates a topology $\mathcal T$ having $\mathcal B$ as a base), no additional arguments are needed to show that the $\varphi_\alpha$ are homeomorphisms.

However, you definitely need more than that. Theoretically it could be that $U_\alpha$ contains elements of $\mathcal B$ not having the form $\varphi_\alpha^{-1}(V)$, and you correctly state as a lemma that this cannot happen. This requires again (ii) and (iii).

On the other hand, we may argue that the lemma is covered by what has been already proved. In fact, if $\varphi_\beta^{-1}(W) \subset U_\alpha$, then (1) shows that $$\varphi_\beta^{-1}(W) = U_\alpha \cap \varphi_\beta^{-1}(W) = \varphi_\alpha^{-1}(\varphi_\alpha(U_\alpha)) \cap \varphi_\beta^{-1}(W) = \varphi_\alpha^{-1}(\varphi_\alpha(U_\alpha) \cap (\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)) \\ = \varphi_\alpha^{-1}((\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)) = \varphi_\alpha^{-1}(V)$$ where $V = (\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\mathbb R^n$.

In that sense we can agree that the $\varphi_\alpha$ are homeomorphisms "essentially by definition".

Remark: Perhaps it is a bit nitpicky, but Lee is not completely precise. He considers $\varphi_\beta\circ\varphi_\alpha^{-1}$ as a map from to $\varphi_\alpha(U_\alpha\cap U_\beta)$ to $\varphi_\beta(U_\alpha\cap U_\beta)$. Doing so, $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is not defined unless $W \subset \varphi_\beta(U_\alpha\cap U_\beta)$. Thus the "correct" approach would be either to replace $W$ by $W \cap \varphi_\beta(U_\alpha\cap U_\beta)$ or to consider $\varphi_\beta\circ\varphi_\alpha^{-1}$ as a map from $\varphi_\alpha(U_\alpha\cap U_\beta)$ to $\mathbb R^n$.

Q2:

You want to work with two types of maps $\varphi_\alpha$. Type one has range $\mathbb R^n$, type two has range $\mathbb H^n$. Let us write neutrally $\varphi_\alpha : U_\alpha \to \mathbb S_\alpha$, where $S_\alpha$ is one of $\mathbb R^n, \mathbb H^n$. Then the requirements (i) and (ii) read as

(i) $\varphi_\alpha(U_\alpha)$ is open in $\mathbb S_\alpha$.

(ii) $\varphi_\alpha(U_\alpha\cap U_\beta)$ is open in $\mathbb S_\alpha$.

By the way, (i) is redundant (in (ii) we can take $\alpha = \beta$). Note that it is essential to require the sets to be open in $\mathbb S_\alpha$ which is the range of $\varphi_\alpha$. You cannot expect that $\varphi_\alpha(U_\alpha\cap U_\beta)$ is open in $\mathbb R^n$ if $\mathbb S_\alpha = \mathbb H^n$. Anyway, by (iii) we get the smooth (in particular continuous) transition maps $$\varphi_\beta\circ\varphi_\alpha^{-1} : \varphi_\alpha(U_\alpha\cap U_\beta) \to \varphi_\beta(U_\alpha\cap U_\beta)$$ or alternatively (see my above remark) $$\varphi_\beta\circ\varphi_\alpha^{-1} : \varphi_\alpha(U_\alpha\cap U_\beta) \to \mathbb S_\beta .$$ For showing that $ \varphi_\alpha^{-1}(V)\cap\varphi_\beta^{-1}(W) \in \mathcal B$, you do not have show that $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\mathbb{R}^n$. In fact, this may be wrong. What you know is that $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\varphi_\alpha(U_\alpha\cap U_\beta)$ which is open in $\mathbb S_\alpha$. Hence $(\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$ is open in $\mathbb S_\alpha$ and and so is $V \cap (\varphi_\beta\circ\varphi_\alpha^{-1})^{-1}(W)$.

Remark: For manifolds with boundary there is no benefit in working with two type of charts (having range $\mathbb R^n$ or $\mathbb H^n$) instead of working only with charts having range $\mathbb H^n$. In fact, charts $\varphi : U \to V \subset \mathbb H^n$ can be divided into two classes: One has the property $\varphi(U) \cap \mathbb R^{n-1} \times \{ 0 \} = \emptyset$ (charts around interior points), the other has the property $\varphi(U) \cap \mathbb R^{n-1} \times \{ 0 \} \ne \emptyset$ (charts around boundary points).