297: Quotient of Cylinder with Antipodal Points Identified Is Homeomorphic to Möbius Band

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A description/proof of that quotient of cylinder with antipodal points identified is homeomorphic to Möbius Band

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note 1
• 4: Getting the Head Around the Correspondences
• 5: Note 2

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Starting Context

• The reader knows a definition of Möbius band.
• The reader knows a definition of quotient topology on set with respect to map.
• The reader admits the universal property of quotient map.

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Target Context

• The reader will have a description and a proof of the proposition that the quotient of the $S^1\times [-1,1]$ cylinder with antipodal points identified is homeomorphic to the Möbius band.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

The quotient of the $S^1\times [-1,1]$ cylinder with antipodal points identified, $(S^1\times [-1,1])/\sim$ where $(e^{\theta i},z)\sim (-e^{\theta i},-z)$ where $-\pi <\theta \le \pi$, is homeomorphic to the Möbius band, $([0,1]\times [0,1])/\sim$ where $(x,0)\sim (1-x,1)$.

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2: Proof

Let us choose the unique representative of each point of $(S^1\times [-1,1])/\sim$. For any $[(e^{\theta i},z)]$ where $z\ne 0$, the unique representative is $0<z$ and $-\pi <\theta \le \pi$. For any $[(e^{\theta i},z)]$ where $z=0$, the unique representative is $z=0$ and $0\le \theta <\pi$. So, we can think only of the $0\le z\le 1$ part of the cylinder with the half of the equator removed. We will always think of each point of the quotient space in term of the unique representative.

Let us choose the unique representative of each point of $([0,1]\times [0,1])/\sim$. For any $[(x,y)]$ where $y=0$ or $y=1$, the unique representative is $y=0$. So, we can think only of the square with the upper side removed. We will always think of each point of the quotient space in term of the unique representative.

Let us think of the map, $f:(S^1\times [-1,1])/\sim \to ([0,1]\times [0,1])/\sim$, $[(e^{\theta i},z)]\mapsto [((1/2)(1+z),0)]$ when $\theta =\pi$ which implies that $0<z\le 1$, $[(e^{\theta i},z)]\mapsto [((1/2)(1-z),\theta /\pi )]$ when $0\le \theta <\pi$ which implies that $0\le z\le 1$, and $[(e^{\theta i},z)]\mapsto [((1/2)(1+z),\theta /\pi +1)]$ when $-\pi <\theta <0$ which implies that $0<z\le 1$. That is well-defined, because as we are thinking in terms of unique representatives, there is no duplication of specifications, and it maps each unique representative to a unique representative.

Let us prove that $f$ is injective. If $[(e^{\theta i},z)]\ne [(e^{{\theta }^{\prime }i},z^{\prime })]$, $\theta \ne {\theta }^{\prime }$ or ($\theta ={\theta }^{\prime }$ and $z\ne z^{\prime }$). Let us suppose that $\theta \ne {\theta }^{\prime }$. Let us suppose ${\theta }^{\prime }<\theta$ without loss of generality. There are the cases, 1): $\theta =\pi$ and $0\le {\theta }^{\prime }<\pi$; 2): $\theta =\pi$ and $-\pi <{\theta }^{\prime }<0$; 3): $0\le \theta <\pi$ and $0\le {\theta }^{\prime }<\pi$; 4): $0\le \theta <\pi$ and $-\pi <{\theta }^{\prime }<0$; 5): $-\pi <\theta <0$ and $-\pi <{\theta }^{\prime }<0$. For the case, 1), as $\theta =\pi$ implies $0<z$, $(1/2)(1+z)\ne (1/2)(1-z^{\prime })$, because if $(1/2)(1+z)=(1/2)(1-z^{\prime })$, $z+z^{\prime }=0$, which would be impossible as $0\le z^{\prime }$. For the case, 2), $0\ne {\theta }^{\prime }/\pi +1$. For the case, 3), $\theta /\pi \ne {\theta }^{\prime }/\pi$. For the case, 4), as $-\pi <{\theta }^{\prime }<0$ implies $0<z^{\prime }$, $(1/2)(1-z)\ne (1/2)(1+z^{\prime })$, because if $(1/2)(1-z)=(1/2)(1+z^{\prime })$, $z+z^{\prime }=0$, which would be impossible as $0\le z$. For the case, 5), $\theta /\pi +1\ne {\theta }^{\prime }/\pi +1$. If ($\theta ={\theta }^{\prime }$ and $z\ne z^{\prime }$), $(1/2)(1+z)\ne (1/2)(1+z^{\prime })$ when $\theta =\pi$ or $-\pi <\theta <0$ and $(1/2)(1-z)\ne (1/2)(1-z^{\prime })$ when $0\le \theta <\pi$.

Let us prove that $f$ is surjective. For any $[(x,y)]\in ([0,1]\times [0,1])/\sim$, there are the cases, 1): $y=0$; 2): $0<y<1$. For the case, 1), $(x,0)$ where $1/2<x\le 1$ is realized by $((1/2)(1+z),0)=(x,0)$ where $\theta =\pi$ and $0<z\le 1$; $(x,0)$ where $0\le x\le 1/2$ is realized by $((1/2)(1-z),\theta /\pi )=(x,0)$ where $\theta =0$ and $0\le z\le 1$. For the case, 2), $(x,y)$ where $0\le x\le 1/2$ is realized by $((1/2)(1-z),\theta /\pi )=(x,y)$ where $0<\theta <\pi$ and $0\le z\le 1$; $(x,y)$ where $1/2<x\le 1$ is realized by $((1/2)(1+z),\theta /\pi +1)=(x,y)$ where $-\pi <\theta <0$ and $0<z\le 1$.

So, $f$ is a bijection.

Let us prove that $f$ is continuous.

There are the quotient maps, $\pi :S^1\times [-1,1]\to (S^1\times [-1,1])/\sim$ and ${\pi }^{\prime }:[0,1]\times [0,1]\to ([0,1]\times [0,1])/\sim$.

Let us define $\tilde{f}:S^1\times [-1,1]\to ([0,1]\times [0,1])/\sim =f\circ \pi$. $f$ is continuous if and only if $\tilde{f}$ is continuous, by the universal property of quotient map.

Let us think of any point, $(e^{\theta i},z)\in S^1\times [-1,1]$, and any open neighborhood of $[(x,y)]=\tilde{f}((e^{\theta i},z))$, $U_{[(x,y)]}$. If $y\ne 0$, there is a smaller open neighborhood, $U_{[(x,y)]}^{\prime }\subseteq U_{[(x,y)]}$, such that ${\pi }^{\prime -1}(U_{[(x,y)]}^{\prime })=B_{(x,y)-ϵ}\cap ([0,1]\times [0,1])$, where $B_{(x,y)-ϵ}$ is an open ball (likewise notations will be used hereafter without further explanations) that does not contain any point whose $y$-component is $0,1$. If $y=0$, there is a smaller open neighborhood, $U_{[(x,y)]}^{\prime }\subseteq U_{[(x,y)]}$, such that ${\pi }^{\prime -1}(U_{[(x,y)]}^{\prime })=(B_{(x,0)-ϵ}\cap ([0,1]\times [0,1]))\cup (B_{(1-x,1)-ϵ}\cap ([0,1]\times [0,1]))$.

If $\theta \ne 0,\pi$, which implies that $y\ne 0$, there is an open neighborhood, $U_{(e^{\theta i},z)}=e^{i((\theta -\delta ,\theta +\delta ))}\times ((z-\delta ,z+\delta )\cap [-1,1])\subseteq S^1\times [-1,1]$, where $e^{i((\theta -\delta ,\theta +\delta ))}$ is the image of $(\theta -\delta ,\theta +\delta )$ under the $e^{i?}$ function (likewise notations will used hereafter without further explanations) such that $0,\pi \notin (\theta -\delta ,\theta +\delta )$. Then, $U_{(e^{\theta i},z)}$ is entirely in the $0\le \theta <\pi$ area or the $-\pi <\theta <0$ area, so, $\tilde{f}((e^{{\theta }^{\prime }i},z^{\prime }))$ is $[((1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi )]$ or $[((1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1)]$ there, and $\delta$ can be chosen to be small enough such that $((1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi )$ or $((1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1)$ is in $B_{(x,y)-ϵ}\cap ([0,1]\times [0,1])$, because $(1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi ,(1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1$ are continuous with respect to ${\theta }^{\prime },z^{\prime }$, so, $\tilde{f}(U_{(e^{\theta i},z)})\subseteq U_{[(x,y)]}^{\prime }$.

If $\theta =0$, which implies that $y=0$, there is an open neighborhood, $U_{(e^{\theta i},z)}=e^{i((-\delta ,\delta ))}\times ((z-\delta ,z+\delta )\cap [-1,1])\subseteq S^1\times [-1,1]$. Then, $U_{(e^{\theta i},z)}$ consists of the $0\le \theta <\pi$ area and the $-\pi <\theta <0$ area, so, $\tilde{f}((e^{{\theta }^{\prime }i},z^{\prime }))$ is $[((1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi )]$ when $0\le {\theta }^{\prime }<\pi$ and $[((1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1)]$ when $-\pi <{\theta }^{\prime }<0$, and $\delta$ can be chosen to be small enough such that $((1/2)(1-z^{\prime }),{\theta }^{\prime }/{\pi }^{\prime })\in (B_{(x,0)-ϵ}\cap ([0,1]\times [0,1]))$ for the former area, because $(1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi$ are continuous with respect to ${\theta }^{\prime },z^{\prime }$ and $((1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1)\in (B_{(1-x,1)-ϵ}\cap ([0,1]\times [0,1]))$ for the latter area, because $(1/2)(1+z)=1-x$ and $0/\pi +1=1$ and $(1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1$ are continuous with respect to ${\theta }^{\prime },z^{\prime }$, so, $\tilde{f}(U_{(e^{\theta i},z)})\subseteq U_{[(x,y)]}^{\prime }$.

If $\theta =\pi$, which implies that $y=0$, there is an open neighborhood, $U_{(e^{\theta i},z)}=e^{i((\pi -\delta ,\pi )\cup \{\pi \}\cup (-\pi ,-\pi +\delta ))}\times ((z-\delta ,z+\delta )\cap [-1,1])\subseteq S^1\times [-1,1]$. Then, $U_{(e^{\theta i},z)}$ consists of the $\theta =\pi$ area, the $0\le \theta <\pi$ area, and the $-\pi <\theta <0$ area, so, $\tilde{f}((e^{{\theta }^{\prime }i},z^{\prime }))$ is $[((1/2)(1+z^{\prime }),0)]$ when ${\theta }^{\prime }=\pi$, $[((1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi )]$ when $0\le {\theta }^{\prime }<\pi$, and $[((1/2)(1+z^{\prime }),0)]$ when $-\pi <{\theta }^{\prime }<0$, and $\delta$ can be chosen to be small enough such that $((1/2)(1+z^{\prime }),0)\in (B_{(x,0)-ϵ}\cap ([0,1]\times [0,1]))$ for the 1st area, because $(1/2)(1+z^{\prime })$ is continuous with respect to $z^{\prime }$, $((1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi )\in (B_{(1-x,1)-ϵ}\cap ([0,1]\times [0,1]))$ for the 2nd area, because $(1/2)(1-z)=1-x$ and $\pi /\pi =1$ and $(1/2)(1-z^{\prime }),{\theta }^{\prime }/\pi$ are continuous with respect to ${\theta }^{\prime },z^{\prime }$, and $((1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1)\in (B_{(x,0)-ϵ}\cap ([0,1]\times [0,1]))$ for the 3rd area, because $(1/2)(1+z^{\prime }),{\theta }^{\prime }/\pi +1$ are continuous with respect to $thet{a}^{\prime },z^{\prime }$, so, $\tilde{f}(U_{(e^{\theta i},z)})\subseteq U_{[(x,y)]}^{\prime }$.

So, $\tilde{f}$ is continuous at any point of $S^1\times [-1,1]$.

$f^{-1}:([0,1]\times [0,1])/\sim \to (S^1\times [-1,1])/\sim$ is $[(x,y)]\mapsto [(e^{\pi yi},1-2x)]$ when $0\le x\le 1/2$, $[(x,y)]\mapsto [(e^{\pi i},2x-1)]$ when $1/2<x\le 1$ and $y=0$, $[(x,y)]\mapsto [(e^{\pi (y-1)i},2x-1)]$ when $1/2<x\le 1$ and $0<y<1$.

Let us define $\widetilde{f^{-1}}:[0,1]\times [0,1]\to (S^1\times [-1,1])/\sim =f^{-1}\circ {\pi }^{\prime }$. $f^{-1}$ is continuous if and only if $\widetilde{f^{-1}}$ is continuous, by the universal property of quotient map.

Let us think of any point, $(x,y)\in [0,1]\times [0,1]$, and any open neighborhood of $[(e^{\theta i},z)]=\widetilde{f^{-1}}((x,y))$, $U_{[(e^{\theta i},z)]}$. If $z\ne 0$, there is a smaller open neighborhood, $U_{[(e^{\theta i},z)]}^{\prime }\subseteq U_{[(e^{\theta i},z)]}$, such that ${\pi }^{-1}(U_{[(e^{\theta i},z)]}^{\prime })=((e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [-1,1]))\cup ((e^{i((\theta -\pi -ϵ,\theta -\pi +ϵ))})\times ((-z-ϵ,-z+ϵ)\cap [-1,1]))$ where $(\theta -ϵ,\theta +ϵ),(\theta -\pi -ϵ,\theta -\pi +ϵ)$ are really translated to be in $(-\pi ,\pi ]$ and $0\notin (z-ϵ,z+ϵ)$. If $z=0$, there is a smaller open neighborhood, $U_{[(e^{\theta i},z)]}^{\prime }\subseteq U_{[(e^{\theta i},z)]}$, such that ${\pi }^{-1}(U_{[(e^{\theta i},z)]}^{\prime })=((e^{i((\theta -ϵ,\theta +ϵ))})\times (-ϵ,ϵ))\cup ((e^{i((\theta -\pi -ϵ,\theta -\pi +ϵ))})\times (-ϵ,ϵ))$ where $(\theta -ϵ,\theta +ϵ),(\theta -\pi -ϵ,\theta -\pi +ϵ)$ are really translated to be in $(-\pi ,\pi ]$.

If $0\le x<1/2$, which implies that $z\ne 0$, there is an open neighborhood, $U_{(x,y)}=B_{(x,y)-\delta }\cap ([0,1]\times [0,1])\subseteq [0,1]\times [0,1]$ such that $1/2\notin (x-\delta ,x+\delta )$. Then, $U_{(x,y)}$ is entirely in the $0\le x\le 1/2$ area, so, $\widetilde{f^{-1}}((x^{\prime },y^{\prime }))$ is $[(e^{\pi y^{\prime }i},1-2x^{\prime })]$ there, and $\delta$ can be chosen to be small enough such that $(e^{\pi y^{\prime }i},1-2x^{\prime })$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [0,1])$, because $e^{\pi y^{\prime }i},1-2x^{\prime }$ are continuous with respect to $x^{\prime },y^{\prime }$, so, $\widetilde{f^{-1}}(U_{(x,y)})\subseteq U_{[(e^{\theta i},z)]}^{\prime }$.

If $1/2<x\le 1$ and $0<y\le 1$, which implies that $z\ne 0$, there is an open neighborhood, $U_{(x,y)}=B_{(x,y)-\delta }\cap ([0,1]\times [0,1])\subseteq [0,1]\times [0,1]$ such that $1/2\notin (x-\delta ,x+\delta )$ and $0\notin (y-\delta ,y+\delta )$. Then, $U_{(x,y)}$ is entirely in the $1/2<x\le 1$ and $0<y\le 1$ area, so, $\widetilde{f^{-1}}((x^{\prime },y^{\prime }))$ is $[(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)]$ there, and $\delta$ can be chosen to be small enough such that $(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [0,1])$, because $e^{\pi (y^{\prime }-1)i},2x^{\prime }-1$ are continuous with respect to $x^{\prime },y^{\prime }$, so, $\widetilde{f^{-1}}(U_{(x,y)})\subseteq U_{[(e^{\theta i},z)]}^{\prime }$.

If $1/2<x\le 1$ and $y=0$, which implies that $z\ne 0$, there is an open neighborhood, $U_{(x,y)}=B_{(x,y)-\delta }\cap ([0,1]\times [0,1])\subseteq [0,1]\times [0,1]$ such that $1/2\notin (x-\delta ,x+\delta )$. Then, $U_{(x,y)}$ consists of the $1/2<x\le 1$ and $y=0$ area and the $1/2<x\le 1$ and $0<y\le 1$ area, so, $\widetilde{f^{-1}}((x^{\prime },y^{\prime }))$ is $[(e^{\pi i},2x^{\prime }-1)]$ when $y^{\prime }=0$ and $[(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)]$ when $0<y^{\prime }\le 1$, and $\delta$ can be chosen to be small enough such that $(e^{\pi i},2x^{\prime }-1)$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [0,1])$, because $2x^{\prime }-1$ is continuous with respect to $x^{\prime }$ and $(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [0,1])$, because $e^{\pi (0-1)i}=e^{-\pi i}=e^{\pi i}$ and $e^{\pi (y^{\prime }-1)i},2x^{\prime }-1$ are continuous with respect to $x^{\prime },y^{\prime }$, so, $\widetilde{f^{-1}}(U_{(x,y)})\subseteq U_{[(e^{\theta i},z)]}^{\prime }$.

If $x=1/2$ and $0<y\le 1$, which implies that $z=0$, there is an open neighborhood, $U_{(x,y)}=B_{(x,y)-\delta }\subseteq [0,1]\times [0,1]$ such that $0\notin (y-\delta ,y+\delta )$. Then, $U_{(x,y)}$ consists of the $0\le x\le 1/2$ area and the $1/2<x\le 1$ and $0<y\le 1$ area, so, $\widetilde{f^{-1}}((x^{\prime },y^{\prime }))$ is $[(e^{\pi yi},1-2x^{\prime })]$ when $x^{\prime }\le 1/2$ and $[(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)]$ when $1/2<x^{\prime }$, and $\delta$ can be chosen to be small enough such that $(e^{\pi y^{\prime }i},1-2x^{\prime })$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times (z-ϵ,z+ϵ)$, because $e^{\pi y^{\prime }i},1-2x^{\prime }$ are continuous with respect to $x^{\prime },y^{\prime }$ and $(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)$ is in $(e^{i((\theta -\pi -ϵ,\theta -\pi +ϵ))})\times ((-z-ϵ,-z+ϵ)\cap [0,1])$, because $e^{\pi (y-1)i}=e^{(\pi y-\pi )i}=e^{(\theta -\pi )i}$ and $2x-1=0=-z$ and $e^{\pi (y^{\prime }-1)i},2x^{\prime }-1$ are continuous with respect to $x^{\prime },y^{\prime }$, so, $\widetilde{f^{-1}}(U_{(x,y)})\subseteq U_{[(e^{\theta i},z)]}^{\prime }$.

If $x=1/2$ and $y=0$, which implies that $z=0$, there is an open neighborhood, $U_{(x,y)}=B_{(x,y)-\delta }\cap ([0,1]\times [0,1])\subseteq [0,1]\times [0,1]$. Then, $U_{(x,y)}$ consists of the $0\le x\le 1/2$ area, the $1/2<x\le 1$ and $y=0$ area, and the $1/2<x\le 1$ and $0<y\le 1$ area, so, $\widetilde{f^{-1}}((x^{\prime },y^{\prime }))$ is $[(e^{\pi y^{\prime }i},1-2x^{\prime })]$ when $x^{\prime }\le 1/2$, $[(e^{\pi i},2x^{\prime }-1)]$ when $1/2<x^{\prime }$ and $y^{\prime }=0$, and $[(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)]$ when $1/2<x^{\prime }$ and $0<y^{\prime }\le 1$, and $\delta$ can be chosen to be small enough such that $(e^{\pi y^{\prime }i},1-2x^{\prime })$ is in $(e^{i((\theta -ϵ,\theta +ϵ))})\times ((z-ϵ,z+ϵ)\cap [0,1])$, because $e^{\pi y^{\prime }i},1-2x^{\prime }$ are continuous with respect to $x^{\prime },y^{\prime }$, $(e^{\pi i},2x^{\prime }-1)$ is in $(e^{i((\theta -\pi -ϵ,\theta -\pi +ϵ))})\times ((-z-ϵ,-z+ϵ)\cap [0,1])$, because $e^{\pi i}=e^{(-\pi )i}=e^{(\theta -\pi )i}$ and $2x-1=0=-z$ and $2x^{\prime }-1$ is continuous with respect to $x^{\prime },y^{\prime }$, and $(e^{\pi (y^{\prime }-1)i},2x^{\prime }-1)$ is in $(e^{i((\theta -\pi -ϵ,\theta -\pi +ϵ))})\times ((-z-ϵ,-z+ϵ)\cap [0,1])$, because $e^{\pi (y-1)i}=e^{(\pi y-\pi )i}=e^{(\theta -\pi )i}$ and $2x-1=0=-z$ and $e^{\pi (y^{\prime }-1)i},2x^{\prime }-1$ are continuous with respect to $x^{\prime },y^{\prime }$, so, $\widetilde{f^{-1}}(U_{(x,y)})\subseteq U_{[(e^{\theta i},z)]}^{\prime }$.

So, $\widetilde{f^{-1}}$ is continuous at any point of $[0,1]\times [0,1]$.

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3: Note 1

A pitfall about the Möbius band or more generally any 2-dimensional topological space embedded in any 3-dimensional Euclidean space is to think that the space has the front and the back, which is not about whether the space is oriented or not but about imagining that there are a point on the front and the corresponding point on the back as 2 different points.

That misunderstanding seems to tend to happen because one imagines a sheet of paper as the space. The sheet of paper has the front and the bottom, and when one draws a curve on the front, the back is blank, and one can draw another curve on the back.

The sheet of paper has the front and the bottom because it is really a 3-dimensional object that has the thickness.

A pitfall when one has a paper model of the Möbius band is to think that when one draws a curve on the Möbius band, the curve returns to the starting point only when the curve returns to the starting point on the sheet of paper; no, when the curve on the sheet goes to the point on the back of the starting point, the curve has already returned to the starting point on the space.

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4: Getting the Head Around the Correspondences

We already know the correspondences of points between the Möbius band and the quotient of the cylinder as formulas, but intuitively, the correspondences are not immediately clear. For example, where a circle on the Möbius band is on the quotient of the cylinder?

We think of the quotient of the cylinder in terms of the unique representations of points cited in Proof, which means that we can think of only the $0\le z\le 1$ part of the cylinder with the half of the equator removed.



The border of the Möbius band is really a single circle, which seems to correspond to the $z=1$ circle of the cylinder, which is really so?



According to the formulas of $f^{-1}$, $x=0$ on the Möbius band corresponds to $(e^{y\pi i},1)$ on the cylinder, which is certainly the half of the $z=1$ circle of the cylinder, and $x=1$ on the Möbius band corresponds to $(e^{(y-1)\pi i},1)$ on the cylinder, which is certainly the remaining half of the $z=1$ circle of the cylinder. So, yes.

Where is the central circle on the Möbius band on the cylinder?


It on the Möbius band is $x=1/2$. Is that so short?, I mean, it seems half of the border. Yes, as noted in Note 1, the central circle has returned to the starting point when the curve has come to the back of the starting point on the paper sheet, not to the starting point on the sheet. So, although it is not closed on the sheet, it is closed on the space.

Anyway, according to the formulas of $f^{-1}$, $x=1/2$ on the Möbius band corresponds to $(e^{y\pi i},0)$ on the cylinder. So, the central circle on the Möbius band is the half circle at $z=0$ on the cylinder, which is really closed on the quotient space, because antipodal points on the circle are identified.


Where is a non-central circle on the Möbius band on the cylinder?

For example, let us think of the $x=1/4,3/4$ circle on the Möbius band. Note that it is about twice-long the central circle, because the non-central circle does not pass the starting point on the back of the sheet but returns to the starting point only on the front of the sheet.


According to the formulas of $f^{-1}$, $x=1/4$ on the Möbius band corresponds to $(e^{y\pi i},1/2)$ on the cylinder, and $x=3/4$ on the Möbius band corresponds to $(e^{(y-1)\pi i},1/2)$ on the cylinder. So, the non-central circle on the Möbius band is the circle at $z=1/2$ on the cylinder.


What if the non-central curve is twisted to return to the starting point on the back of the sheet?

Let us think of a line segment that is perpendicular to the central circle, which (the line segment) connects 2 points on the border, on the Möbius band (let us call the line segment 'cross line segment').

For example, it is $y=1/2$.


According to the formulas of $f^{-1}$, $y=1/2$ on the Möbius band corresponds to $(e^{(1/2)\pi i},1-2x)$ where $0\le x\le 1/2$ and $(e^{-(1/2)\pi i},2x-1)$ where $1/2<x\le 1$ on the cylinder So, the cross line segment on the Möbius band starts at a border point, goes perpendicularly down to $z=0$, jumps to the antipodal point of the $z=0$ circle, and goes perpendicularly up to the point on the border that is antipodal to the starting point on the border circle, on the cylinder.


In order to twist the non-central curve to return to the starting point on the back of the sheet, we can take a half circle of $z=1/2$, and connect the terminal of the half circle to the starting point via the part of the cross line segment.



Let us see that the quotient space of the cylinder is not really oriented.

Let us think of the central circle on the Möbius band. Let us have a basis at the starting point ($\theta =0$) of the $z=0$ half circle on the cylinder with the 1st vector as along the half circle and the 2nd vector as parallel to the z axis. Then, at the terminal point ($\theta =\pi$), which is really the same point with the starting point on the quotient space, the 1st vector reflects back to the same vector at the starting point and the 2nd vector to the opposite vector at the starting point.




Note that for any non-central circle on the Möbius band, the basis returns to the original basis, which is not odd at all, because that amounts to returning to the starting point on the front on the paper sheet. In order to close the non-central curve on the back on the paper sheet, the curve has to be twisted, which means that the curve crosses the $z=0$ circle, which makes the basis reversed.

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5: Note 2

When we think of any curve that connects a point on the border to another point on the border, there are 2 types: 1) the curve encloses some points; 2) the curve encloses no point. "enclose" means that some 2 points on the space cannot be path-connected with a path that does not cross the curve.

This is an example that encloses some points.


Note that that is not because the 2 points are on the same $x=0$ part: also this encloses some points.


This is an example that encloses no point.


That seems obvious for the straightforward (non-Möbius) band. The straightforward band has 2 borders, and an intuition is that a curve that connects 2 points on the same border encloses some points while a curve that connects 2 points on the different borders encloses no point.

But the Möbius band has only 1 border and curves have no distinction in that point, but still, some curves enclose some points and the other curves enclose no point.

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References

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