# Break up $\mathbb{R}P^2$ into a part homeomorphic to Mobius band & part homeo. to the 2-disc

The claim is that $\mathbb{R}P^2 = A \cup B$ where $A \simeq$ Mobius band, $B \simeq D^2$, and $A \cap B \simeq S^1$. I understand this intuitively with a gluing type argument, similar to the arguments here https://math.stackexchange.com/questions/77569/bbb-rp2-as-the-union-of-a-möbius-band-and-a-disc, stating generally that if you take a disc away from the projective plane the result should be a mobius strip.

But I'm wondering if there is a way to explicitly write things down, giving rules for homeomorphisms, and if anyone has a hint for how I can start thinking about this in a more precise way.

Start with the 2--1 covering map $$p : S^2 \to \mathbb{R} P^2$$ which is a quotient map having the property that $p(x)=p(y) \iff x = \pm y$.
Using spherical coordinates on $S^2$ with $\phi$ being latitudinal angle measurement from the north pole $(0,0,1)$, decompose $$S^2 = D_+ \cup C \cup D_-$$ where $D_+$ is the northern polar cap defined by $0 \le \phi \le \pi/4$, $D_-$ is the southern polar cap defined by $3\pi/4 \le \phi \le \pi$, and $C$ is the equatorial strip defined by $\pi/4 \le \phi \le 3 \pi/4$. The set $D_+ \cup D_-$ is saturated with respect to $p$, and the restriction $p \mid D_+ \cup D_-$ is a 2--1 covering map over a subset $B \subset \mathbb{R} P^2$ homeomorphic to a disc. The set $C$ is also saturated, and the restriction $p \mid C$ is a 2--1 covering map over a subset $A$ of $\mathbb{R} P^2$ homeomorphic to the Mobius band. Note that the intersection of $\partial C$ and $\partial(D_+ \cup D_-)$ is two circles, and the restriction of $p$ to those circles is a double covering map over a single circle which is equal to both $\partial A$ and $\partial B$. It follows that $\mathbb{R} P^2$ is homeomorphic to the quotient of $A$ and $B$ by identifying their boundaries via a homeomorphism of circles.
• Ah, thanks so much. Just to probe a little bit further - filling in the details in my mind would be writing down universal property diagrams for A and B with the quotient maps given by the mentioned restrictions and proving homeos - is this the right thinking? Also, your last sentence "It follows that ℝP2 is homeomorphic to the quotient of A and B by identifying their boundaries via a homeomorphism of circles." Aren't we just trying to show that $P^2$ will be the explicit union of these sets? I'm not sure where this homeomorphism comes in – Schwinger Oct 17 '14 at 2:49
• @Schwinger: about "universal properties", yes, that's a good catchall way to refer to the various theorems I mentioned at the end. About the "explicit union", the quotient point of view is more powerful, in that it generalizes: you can think of the mobius band and the disc as being a disjoint union, and then produce $\mathbb{R}P^2$ as a quotient as said in my answer. While it is not necessary to do this in the situation of your question, I guess I was thinking more generally where it is necessary, e.g. when thinking of the torus as a square with opposite sides glued. – Lee Mosher Oct 17 '14 at 13:08