Do These Charts Define an Atlas on M?

Question

Question:

Let $M\subseteq \mathbb{R}^2$ be the boundary of the square with vertices at $(\pm 1,\pm 1)$:

$M = \{(x,y): |x|\leq 1$ and $|y|\leq 1$, with $|x|=1$ or $|y|=1\}$.

Decide whether or not the charts $(U,\phi)$ and $(V,\psi)$ with

$U := \{(x,y)\in M: y>-1\}, \phi (x,y) = \frac{x}{1+y}$

$V := \{(x,y)\in M: y<1\}, \phi (x,y) = \frac{x}{1-y}$

define an atlas on M. Prove your answer.

My struggles:

(1) I think the charts do not cover $M$ since $U\cup V$ do not account for the specific conditions $|x|\leq 1$ and $|y|\leq 1$ and $|x|=1$ and $|y|=1$. However, I am not so confident about this.

(2) In case the charts define an atlas on M, I don't know how to come up with the transition map $\psi \circ \phi ^{-1}$ since I don't seem to be able to find $\phi ^{-1}$.

I will really appreciate your help. Thanks in advance.

Answer 1

$U$ and $V$ do cover $M$, as either $y>-1$ or $y<1$ (or both). Next you have to check that the map defines a bijection to $\mathbb{R}$. I highly recommend drawing a picture. Let us consider the case of $U$, then the line connecting $(-1,-1)$ to $(-1,1)$ gets mapped to $(-\infty, -\frac{1}{2}]$, the line connecting $(-1,1)$ to $(1,1)$ gets mapped to $[-\frac{1}{2}, \frac{1}{2}]$ and the line conecting $(1,1)$ to $(1,-1)$ gets mapped to $[\frac{1}{2}, -\infty)$, so this is definitely a bijection. The case with $V$ is up to a change of sign morally the same.

If this is an atlas depends on the setting you work in, if you only care about topological manifolds, then the answer is yes, if you care about smooth manifolds, so you want the maps to be diffeomorphisms (which I assume because of the tags that you chose), then the answer is definitely no, as because of the corners, the inverse map is not smooth.

EDIT: As I was asked to make the last paragraph more explicit, here we are:

We will write down the equation of the inverse for the three lines:

On the line between $(-1,-1)$ and $(-1,1)$ the variable $x$ is constant, so here the the chart is given by $\varphi(y)=\frac{-1}{1+y}$, so the inverse is given by \begin{align*} \psi_1: (-\infty, -\frac{1}{2}] &\to M \\ z &\mapsto (-1, \frac{-1}{z}-1) \end{align*}

On the line from $(-1,1)$ to $(1,1)$ the variable $y$ is constant, so here the chart is $\varphi(x)=\frac{x}{2}$, so the inverse is given by

\begin{align*} \psi_2: [-\frac{1}{2}, \frac{1}{2}] &\to M \\ z &\mapsto (2z, 1) \end{align*}

and for the same reason as in the first case we have \begin{align*} \psi_3: [\frac{1}{2} ,-\infty) &\to M \\ z &\to (1, \frac{1}{z}-1) \end{align*}

Now you can compute the transition function.