# 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.

## 1 Answer

The interesting thing about these gluing arguments is that they are simultaneously intuitive and rigorous. To make them rigorous, to describe the homeomorphisms explicitly, you use quotient maps. Here's how to do it for your question.

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.

The tools needed to fill in some of the rigorous details are various theorems such as can be found for example in Munkres topology, for example: gluing theorems of continuous maps; theorems about induced quotient maps; and theorems characterizing homeomorphisms.

• 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