# Real projective plane as an identification space of the Möbius strip.

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.

• Jun 5 at 19:34

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 $$(M\setminus\partial{M})^+\cong(\mathbb RP^2\setminus )^+\cong RP^2$$ 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, $$re^{it}\mapsto \frac{r+1}{2}e^{it}.$$

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