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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.

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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, $re^{it}\mapsto \frac{r+1}{2}e^{it}.$

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  • $\begingroup$ $D^2\setminus\{0\}$ is non-compact, whereas $A$ is compact... $\endgroup$
    – Kevin.S
    Jun 7 at 12:02
  • $\begingroup$ Here , I am considering the annulus removing the inner circle, I should have mentioned it, I agree@Kevin.S $\endgroup$
    – Dey
    Jun 7 at 12:14
  • $\begingroup$ What is your $A\ $? Circle with a disk removed? If it is the case then your homeomorphism can never be possible. $\endgroup$ Jun 7 at 15:45
  • $\begingroup$ You can check now @Phi beta kappa $\endgroup$
    – Dey
    Jun 7 at 16:22
  • $\begingroup$ I think you mean inner disk. Whatever, I don't understand why $$A / \sim\ \cong M \setminus \partial M \cong \mathbb R P^2 \setminus \{[0]\}.$$ $\endgroup$ Jun 7 at 16:52

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