Consider $X:=[0,1] \times (0,1)$, $\sim$ the equivalence relation such that $(0,t) \sim (1,1-t)$, $M:= X/\sim$ the Möbius strip and $\pi:X \hookrightarrow M, \pi (x,y) = [(x,y)]$.
I need a differentiable atlas for the $M$, but I am not getting it.
I have some ideas but anything seems to work.
First, I thought that we can take $\{(\varphi_k,\pi(U_k))\}_{k\in\{1,2\}}$ where $U_1 = [0,1) \times (0,1)$ and $U_2 = (0,1] \times (0,1)$ and $\varphi_k=(\pi|_{U_k})^{-1}$ but the thing is that $\pi(U_k)=M$ by the definition of $\sim$ and I am not sure if in an atlas the open sets have to be different.
Also, I thought that we can take $U_1:=[0,3/4)\times (0,1)$ and $U_2:=(1/4,1] \times (0,1)$ to avoid that the problem of $\pi(U_k)=M$ but there the problem is that $\pi(U_k)$ is not open since $U_k \neq \pi^{-1}(\pi(U_k))$.
So, any help would be appreciated.
Let $U_1$ be the strip minus the vertical line segment $\{0\}\times (0,1) = \{1\}\times (0,1)$, and $U_2$ be the strip minus the vertical line segment $\{\frac12\}\times (0,1)$. That makes each of the $U_k$ open (since the line segment is closed), and each of them homeomorphic to $(0,1)\times (0,1)$: $U_1$ is actually equal to that space, while $U_2$ needs to be mapped, something like $f: U_2\to (0,1)^2$ given by $$ f(x, y) = \cases{(x + \frac12, y) & if $x< \frac12$\\ (x - \frac12, 1-y) & if $x > \frac12$} $$ should work. You may verify that this respects the relation $\sim$.
Intuitively, this corresponds to cutting across the Möbius strip in two different places, making it into two different rectangles. See below for a concretisation:
Mobius strip with the first vertical line segment marked (note it is exactly on the seam where the paper is taped together):
$U_1$:
Mobius strip with the second vertical line segment marked (note it is nowhere close to the tape):
$U_2$ (note the tape and seam is in the middle of the recangle):