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