# Mobius band is differentiable manifold

Let $X:=[0,1]\times\mathbb{R}$ be equipped with the subspace topology of $\mathbb{R}^2$ and let $(0,t)\sim(1,-t)$ be an equivalence relation on $X$. I must show that the quotient $M:=X/\sim$ equipped with the quotient topology is a differentiable manifold.

I understand that each equivalence class consists of two points:

$[(x,y)]=\{(x,y);(1-x,-y)\}$, i.e. each pair of points corresponding to a reflection at $(\frac{1}{2},0)$.

I first want to show that $M$ is indeed a topological manifold. It is non-empty and a topological space, as is evident. I want to prove that $M$ is hausdorff by showing that $M$ is homeomorphic to $M_{\geq0}:=[0,1]\times\mathbb{R}_{\geq0}$. The space $M_{\geq0}$ is itself hausdorff as a subspace of $\mathbb{R}^2$, if we equip it with the subspace topology.

We can construct the following map.

$\rho:M\rightarrow M_{\geq0}; [(x,y)]\mapsto\begin{cases}(x,y),\quad y\geq0\\ (1-x,-y),\quad y<0\end{cases}$

It is injective:

$\rho([(x,y)])=\rho([(u,v)])\Rightarrow (x,y)=(u,v)\vee(1-x,-y)=(1-u,-v)$,

from which both follows that $(x,y)=(u,v)$.

It is surjective:

Let $x\in[0,1]$ and $y\in\mathbb{R}$.

In the case $y\geq 0$, we obtain $\rho([(x,y)])=(x,y)\in[0,1]\times\mathbb{R}_{\geq0}=M_{\geq0}$.

In the case $y<0$, we obtain $\rho([(x,y)])=(1-x,-y)$,

where $1-x\in[0,1]$ and $-y\in\mathbb{R}_{\geq0}$, hence $\rho([(x,y)])\in[0,1]\times\mathbb{R}_{\geq0}=M_{\geq0}$.

Thus, in all we have $\rho(M)=M_{\geq0}$ and we have shown that $\rho$ is bijective.

Now, how do I show that $\rho$ is continuous? I tried to think of what the open sets from the quotient topology of $M$ look like but that seems to be too complicated an approach because ther are so many cases to be considered because of the subspace topology on $X$ and the corresponding different forms of open sets in $M$. Where is the error in my thoughts?

• $M$ won't be homeomorphic to $M_{\geq 0}$ because the former is nonorientable and the latter is orientable. Your specific $\rho$ has no hope of being a homeomorphism because removing the line $\{x=1/2\}$ from $M$ leaves it connected whereas removing its image under $\rho$, namely $\{x=1/2\}$, from $M_{\geq 0}$, disconnects it. May 12 '18 at 0:03
• You could show that $M$ is Hausdorff directly, by looking at small separating neighborhoods of the preimages of any two points of $M$ in $X$ (this can be done because $X$ is Hausdorff), and pushing the neighborhoods back down into $M$. May 12 '18 at 0:16
• Okay, so what I managed to do is looking at any two $x,y\in X^\circ$ with $x\neq y$ and push their separating neighbourhoods into $M$. This of course yields, that their images are again separating open neighbourhoods in $M$, while $x,y$ are mapped onto themselves. Now I must look at $x,y\in\partial X$. I write their neighbourhoods as the union of their intersection with the interior and the intersection with the boundary: $U=U\setminus\partial X\cup U\cap\partial X$. We then have $\pi(U\setminus\partial X)=U\setminus\partial X$ but $\pi(U\cap\partial X)=$??? May 14 '18 at 13:21
• Did your definition of $M$ contain a typo? Do you want to identify only the boundary points (yielding the classical Möbius strip)? May 14 '18 at 14:36
• Yes, I misunderstood the exercise. The relation was identifying the boundaries with their reflected versions and anything in-between with itself. I'll edit the post. May 14 '18 at 14:49

The idea is to use separating neighbourhoods for representatives of classes $[x] \neq [y]$ in $M$. We can split up the proof in three case: a) $x,y$ both lie in the interior of $X$; b) One of $x$,$y$ lies in the interior of $X$ and the other lies on the boundary; c) Both $x$ and $y$ lie on the boundary of $X$.

a) In this case, pick separating neighbourhoods $U,V$ for $x,y$ respectively that are small enough that they don't touch the boundary. Then since the equivalence relation leaves the interior of $X$ untouched, $\pi(U) \cap \pi(V) = \pi(U \cap V) = \emptyset$.

b) Without loss of generality, we may assume that $x \in \partial X$ and $y \not\in \partial X$. We may find an $\epsilon > 0$ small enough so that $y$ is not contained in the fattened boundary $U = [0,\epsilon]\times \mathbb R \cup [1-\epsilon,1] \times \mathbb R$. Since the complement of $U$ is open in $X$, we may then find a neighborhood $V$ of $y$ that also doesn't intersect $U$. The equivalence relation will leave $V$ intact and perturb the part of $U$ that lies on the boundary, and overall we will have \begin{align}\pi(U) \cap \pi(V) &= (\pi(\partial X) \cup \pi(U \setminus \partial X)) \cap \pi(V) \\&= (\pi(\partial X) \cap \pi(V)) \cup (\pi(U \setminus \partial X) \cap \pi(V))\\ &= \emptyset \cup \pi((U \setminus \partial X) \cap V)\\&= \emptyset.\end{align}

c) If $x,y \in \partial X$, we have to make sure that the separating neighbourhoods in $X$ are small enough so that they don't get squished together in $M$. For example, a tall and skinny open neighbourhood of $(0,1)$ that contains $(0,-1)$ will, down in $M$, intersect a tall and skinny open neighbourhood of $(1,1)$ that contains $(1,-1)$, even though they don't intersect in $X$. The idea, then, is to make sure that the neighbourhoods are not too tall. To this end, choose the representatives $x,y$ so that their first coordinate is $0$, i.e., they both lie on the left boundary of $X$. Then any $U,V$ that separate $x,y$ and don't intersect the right boundary of $X$ will do the job, because the parts of $U,V$ that don't intersect the left boundary boundary will be unchanged by the projection, and the parts that intersect the left boundary won't get mixed up by the projection (as $\pi$ is injective when restricted to $X$ minus the right boundary). I'll leave it to you to write this down as formally as you want to.

• Perfect, that's just what I was thinking. Shrink the neighbourhoods so much that the intersections with the boundaries do not intersect with each other or one of them with the reflection of the other. May 14 '18 at 17:25
• @ThomasWening Yes, exactly! My answer was a bit lengthy, but this same idea can be condensed and reapplied in many other occasions. (For example, this is how one can easily see that projective space is Hausdorff.) To sum things up: pull back to find separating neighbourhoods on the space above, then push back down while making sure the separating neighbourhoods don't contain more than a single representative. May 14 '18 at 17:29

Firstly the mobius strip $M$ can be defined as the quotient space $M = ([0, 1] \times [0, 1])/\sim$ generated by the relation $(x, 0) \sim (1-x, 1)$ for $0 \leq x \leq 1$.

From your question I take it that you want to prove that $M$ is Hausdorff. There is a much simpler way to do this then just by brute-force proving it by the definition of what it means for a topological space to be Hausdorff. (But credits to you for attempting to prove it this way anyhow)

Theorem: The quotient space of a compact Hausdorff space is Hausdorff.

Note now that $[0,1]$ is compact and also Hausdorff, so the product $[0, 1] \times [0,1]$ is compact and also Hausdorff. Since $M$ is the quotient space of $[0, 1] \times [0, 1]$ under a specific relatation by the theorem above we have that $M$ is Hausdorff.

• Your $M$ differs significantly from OP's $M$; his equivalence relation is defined on all $X$, not just the boundary. Also, the theorem as stated is false. You also need the quotient map to be closed. May 12 '18 at 0:14
• Ahh, now I understand! The equivalence relation is only defined on the boundary! Okay, that makes sense. Our sheet wasn't quite clear on that. May 12 '18 at 8:28
• Then again, I do not understand how the equivalence relation works for points $(x,y)$ with $x\notin\{0,1\}$? May 12 '18 at 8:43
• I understand it now: You identify each point in the interior of $X$ with itself and each point on the boundary with iself and its reflection at the point $(1/2,0)$. May 14 '18 at 15:04

So, I wanted to show that $M$ is hausdorff. The approach with the homeomorphism did not help. After some meddling around I came to the solution.

I pick $[x],[y]\in M$ with $[x]\neq[y]$. First, I have to think of what preimages in $X$ fulfill this requirement. One finds that

$$[x]\neq[y]\Rightarrow\begin{cases}x\neq y\wedge x\neq\bar{y}\quad x,y\in\partial X\\ x\neq y,\quad else.\end{cases}$$ where $\bar{x}=\overline{(x_1,x_2)}:=(1-x_1,-x_2)$ is the reflection of $x$ at $(1/2,0)$.

Now for each case we pick such corresponding $x,y\in X$. Then we have $[x]\neq[y]$. Since, in either case we have $x\neq y$, we can find separating open neighbourhoods $U(x), V(y)$ with $U\cap V=\emptyset$. From this follows that $\pi(U\cap V)=\pi(U)\cap\pi(V)=\emptyset$. Because $M$ is equipped with the quotient topology and $U,V\subset X$ are open and $[x]\in\pi(U),[y]\in\pi(V)$, we have separating open neighbourhoods to each $[x]\neq[y]$.

Hence, $M$ is hausdorff, q.e.d..

• You have to be slightly careful. It might happen that separating neighbourhoods for $x,y$ project onto neighbourhoods of $[x],[y]$ that intersect. (The quotient of a Hausdorff space need not be Hausdorff in general.) The error might stem from your assumption $\pi(U \cap V) = \pi(U) \cap \pi(V)$, which does not hold. You need an analysis specific to the problem at hand. I can elaborate in an answer, if you want. May 14 '18 at 16:47
• That would be of great help, thank you. May 14 '18 at 16:54
• I have just thought about it. The only parts of such neighbourhoods in $X$, that may intersect in $M$ are the intersections with $\partial X$, right? My intuition tells me that we can choose them so small in $X$, that they do not intersect anymore in $M$. May 14 '18 at 16:59
• So, from $\pi(U\cap V)=\emptyset$ it follows that $U\cap V\cap X^\circ=\emptyset$ and $U\cap V\cap \partial X=\emptyset$ and $\bar{U}\cap V\cap \partial X=\emptyset$, right? With $\bar{U}=\{\bar{x}|x\in U\}$. May 14 '18 at 17:20