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Prove that $M/\partial M$ is homeomorphic to $\mathbb RP^2,$ where $\partial M$ is the boundary circle of $M.$

My Attempt $:$ Let me first add a diagram here.

enter image description here

The above diagram enables me to write down $\mathbb RP^2$ as a pushout of the following diagram $:$

$$\require{AMScd} \begin{CD}S^1 @>>> D^2 \\ @VVV @VVV \\ M @>>> \mathbb{RP^2}\end{CD}$$

Hence $\mathbb R P^2 \cong M \cup_{\partial} \mathscr D,$ where $\mathscr D$ is the homeomorphic copy of $D^2$ sitting inside $\mathbb {RP}^2.$ Now if we quotient out $\mathscr D$ from $M \cup_{\partial} \mathscr D$ then we get $M/\partial M.$ So we get a quotient map $q : \mathbb {RP^2} \longrightarrow M/\partial M.$ Will it give a homeomorphism?

Any help in this regard will be greatly appreciated. Thanks in advance.

Anil Bagchi.
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1 Answers1

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First prove that, $D^2\setminus\{0\}\cong A.$

Now, $D^2\{0\}/\sim~~\cong A/\sim~~\cong M\setminus\partial{M}$

$M$ being an compact space and $\partial{M}$ is a closed subspace of $M$ ,so $(M\setminus\partial{M})^{+}$ is homeomorphic to $M/\partial{M}$ and $(M\setminus\partial{M})^+\cong(\mathbb RP^2\setminus [0])^+\cong RP^2$ and thus we are done.

Edit (Clarification): $A$ is an half open annulus i.e the inner circle is removed and the desired homeomorphism from $D^2\setminus\{0\}$ to $A$ is given by, $re^{it}\mapsto \frac{r+1}{2}e^{it}.$

SOUL
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  • $D^2\setminus{0}$ is non-compact, whereas $A$ is compact... – Kevin.S Jun 07 '21 at 12:02
  • Here , I am considering the annulus removing the inner circle, I should have mentioned it, I [email protected] – SOUL Jun 07 '21 at 12:14
  • What is your $A\ $? Circle with a disk removed? If it is the case then your homeomorphism can never be possible. – Anil Bagchi. Jun 07 '21 at 15:45
  • You can check now @Phi beta kappa – SOUL Jun 07 '21 at 16:22
  • I think you mean inner disk. Whatever, I don't understand why $$A / \sim\ \cong M \setminus \partial M \cong \mathbb R P^2 \setminus {[0]}.$$ – Anil Bagchi. Jun 07 '21 at 16:52
  • Your idea is intuitively clear to me. Möbius strip is homeomorphic to the antipodal identification space of $D^2$ with the interior of the disk removed from the middle. So if we remove the inner circle from that identification space then we have to remove it's homeomorphic copy in the Möbius strip which is precisely the boundary circle of the Möbius strip. Again you gave an explicit homeomorphism from $D^2 \setminus {0}$ onto $A.$ In the light of it we can conclude that Möbius strip with the boundary circle removed is homeomorphic to $D^2 \setminus {0}$ subject to antipodal identification. – Anil Bagchi. Jun 07 '21 at 19:58
  • Since the interior of $D^2$ goes homeomorphically into the real projective plane $\mathbb R P^2$ it follows that $D^2 \setminus {0}/ x \sim -x \cong \mathbb R P^2 \setminus {[0]}$ and now you combine all these things together to get the desired conclusion. This is essentially what you did, I believe. – Anil Bagchi. Jun 07 '21 at 20:01
  • Yes, you have got it correctly. – SOUL Jun 08 '21 at 02:14
  • Is the answer fine to you ? @Phi beta kappa – SOUL Jun 08 '21 at 11:53