Differentiable atlas for infinite Mobius strip I want to construct a differentiable atlas for the abstract Mobius strip $M:=X/\sim$ equipped with the quotient topology, where $X:=[0,1]\times\mathbb{R}\subset\mathbb{R}^2$ is a subset equipped with the subspace topology and $\sim$ is the equivalence relation, that identifies any $x\in X^\circ$ with itself and any $x\in\partial X$ with itself and its reflection $\bar{x}=\overline{(x_1,x_2)}:=(1-x_1,-x_2)$. I have already shown that $M$ is a Hausdorff space. It is obvious, that it is a non-empty topological space.

1.) Is there any way to show that $M$ is second-countable? My initial idea was to show that $M$ has a global chart. A global chart is a homeomorphism $M\cong U\subset\mathbb{R}^n$. Because second-countability is a topological invariant this would mean that $M$ is second-countable because $U\subset\mathbb{R}^n$ with the subset topology is itself second-countable. But finding such a global chart seems to be impossible.

Let $\pi:X\rightarrow M, x\mapsto [x]$ be the quotient map. Given $\Box:=[0,1]\times (0,1)\subset X$ and 
$$\Phi:\pi(\Box)\rightarrow\Psi\circ\pi(\Box),\quad(\varphi,r)\mapsto(x,y,z),$$
where $$
x:=\cos 2\pi\varphi\left(1+\frac{r}{2}\cos\pi\varphi\right),\\
y:=\sin 2\pi\varphi\left(1+\frac{r}{2}\cos\pi\varphi\right),\\
z:=\frac{r}{2}\sin\pi\varphi,$$
I want to show that $\Phi$ is a chart of $\pi(\Box)\subset M$.
That means I have to show that $\Phi$ is a continuous bijection with an inverse map $\Phi^{-1}$ that is also continuous.
My thoughts on this are the following. To show that $\Phi$ is continuous I want to use that compositions of continuous maps are continuous themselves. However, there is my second problem. 

2.) Since we are dealing with topological spaces, it only makes sense to speak of continuity with regards to the topologies at hand. Does this mean I have to check continuity for all elemental functions contained in $\Phi$?

I have found the following candidate as an inverse function:
$$g: \Phi\circ\pi(\Box)\rightarrow\pi(\Box),\quad(x,y,z)\mapsto(\varphi,r),$$
where $$r:=2\gamma, \quad\varphi:=\frac{1}{\pi}\arccos\frac{R-1}{\gamma},\\
R:=\sqrt{x^2+y^2}, \quad\gamma:=\sqrt{z^2+\left(R-1\right)^2}.$$

3.) Now, to show that, in fact, $g=\Phi^{-1}$, I must show that $\Phi\circ g=id_{\Phi\circ\pi(\Box)}$ and $g\circ\Phi=id_{\pi(\Box)}$. However, upon trying to compute both equalities I never reobtain the input. Is there maybe another way to show that $g=\Phi^{-1}$?

After having shown that $\Psi$ is actually a chart for $\pi(\Box)\subset M$, I want to look at shifted segments $\Box_k:=[0,1]\times(\frac{k}{2},1+\frac{k}{2})$ for $k\in\mathbb{Z}$. Then we obviously have $\Box=\Box_0$ and each adjacent box intersects: $\Box_k\cap\Box_{k+1}=[0,1]\times[\frac{k+1}{2},1+\frac{k}{2}]\neq\emptyset$. It also obviously follows that $\bigcup_{k\in\mathbb{Z}}\Box_k=X$.

4.) The next step would be to define analogous charts
$$\Phi_k:\pi(\Box_k)\rightarrow\Psi\circ\pi(\Box_k),\quad(\varphi,r)\mapsto(x,y,z)$$
for each neighbourhood $\pi(\Box_k)$. And prove that the transition $\tau_{k,k+1}:=\Phi_{k+1}\circ\Phi^{-1}_k$ is differentiable. Since each $\Phi_k$ is differentiable as a composition of differentiable maps,  $\tau_{k,k+1}$ is itself differentiable.
But how do I argue that $\Phi^{-1}_k$ is differentiable?

Given all of the steps above I would then argue that we have an atlas $A_0:=\{(\pi(\Box_k),\Phi_k)|k\in\mathbb{Z}\}$ and therefore a differentiable structure $A=[A_0]$ on $M$. Hence, it is a differentiable manifold.
I am sorry to ask so many questions at once. They all belong to that one exercise that I have been pondering about for the past five days and while I cannot spend much more time pondering so intensely about it, because I have new exercises waiting to be solved, maybe someone can give me some hints, tips or even propose a solution to one of the questions. I find this exercise really good since it involves really revising the definitions and putting the theorems to work.
 A: You are working too hard. Since $M$ is given as a quotient of a nice, rectangular subset of of the plane, we don't need to introduce trigonometric functions. Think of charts as ways of flattening subsets of manifolds. It is intuitively clear that we only need two charts to flatten the whole Möbius strip: cut across the strip in two places, and you are left with two rectangular pieces which you can lay down on the plane. Formally, let $U = \pi(X \setminus \partial X)$, and let $V = \pi(X \setminus \{ x  = 1/2 \})$. Define a chart $\phi:U \to \mathbb R^2$ via $$\phi([x,y]) = (x,y),$$ and a chart $\psi:V \to \mathbb R^2$ via $$\psi([x,y]) = \begin{cases} (x,y) &\text{if } x < 1/2  \\ (x-1,-y) &\text{if } x > 1/2 \end{cases}.$$
You can check that these are well-defined homeomorphisms onto their images, with $\phi^{-1}(x,y) = [x,y]$ and $\psi^{-1}(x,y) = [x,y]$. Moreover, $$\phi \psi^{-1}(x,y) = \phi([x,y]) = (x,y)$$ wherever it is defined and $$\psi \phi^{-1}(x,y) = \psi([x,y]) = \begin{cases} (x,y) &\text{if } 0 < x < 1/2  \\ (x-1,-y) &\text{if } 1 > x > 1/2 \end{cases}; $$ both of these maps are obviously smooth on their respective domains.

Regarding second-countability, we can take open balls or rectangles around points $[x,y] \in M$ with rational $x,y$ as a countable base.
