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This is exercise 6-4 on page 181 in John Lee's Topological Manifolds book which asks me to prove the above, that is, if $M$ is a boundaryless surface which contains a subset $B$ which is homeomorphic to the Möbius band, then $M$ is of the form $M'\#\mathbb RP^2$ where $M'$ is a boundaryless surface.

I tried to attack it in this way:

Try to find a polygonal presentation for $M$ which has a (possibly non-adjacent) twisted pair $aa$. I know how to make it adjacent by means of cutting and pasting (this process is outlined in part I of the proof of the classification theorem). Then we have $M$ presented as $Vaa$ for some word $V$. So set $M'$ to be the surface presented by word $V$ and we are done because concatenation of words gives the presentation of the connected sum and $aa$ presents $\mathbb RP^2$.

The non-trivial part is to find that crucial twisted pair. For this I thought, first triangulate $M$. Then cut along $B$, a Möbius band sitting in $M$. Do we get a polygonal presentation of $M\setminus B$?

Later I looked at the hint which asks us to show that there is $B_0\subsetneq B$ which is homeomorphic to the Möbius band. That's clear to me. But why consider this subset at all?

I know this counts as two questions, but the aim is to just solve this problem.

Thanks in advance!

tomasz
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Karthik C
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1 Answers1

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The answer was almost given by amomin, let us just formulate it precisely.

Take $B_0\subset B$, homeomorphic to the Möbius Band, that you have seen.

Glueing a disk to $M\setminus B_0$ yields a boundaryless surface $M'$. Then, $B_0$ is homeomorphic to $\mathbb{R}P^2$ minus a disc (see $\Bbb RP^2$ as the union of a Möbius band and a disc ).

Removing the disk to $M'$, then glueing $\mathbb{R}P^2$ minus a disc corresponds exactly to do $M' \# \mathbb{R}P^2$, and the result is $M$.

Community
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Jérémy Blanc
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