# Fundamental group of the real projective plane and its universal cover

I have some questions regarding the real projective plane $$\mathbb{R}\mathbb{P}^{2}$$.

If we choose to represent it using the following identification on the square $$[0, 1]^{2}$$ : how can we see that $$\pi_{1}(\mathbb{R}\mathbb{P}^{2})$$ is isomorphic to $$\mathbb{Z} \backslash 2\mathbb{Z}$$ ?

I tried using Seifert-van Kampen theorem but I have a problem. I chose $$U_{1}$$ as the whole square (above) minus the central point and $$U_{2}$$ as a disk included in the square.

First, $$\pi_{1}(U_{2})$$ is trivial (as $$U_{2}$$ is contractible).

Second, the boundary of the square is a deformation retract of $$U_{1}$$ and here is my first question. If we do the same exercise with the Klein bottle or the torus, we know that the boundary is (is homeomorphic ?) a bouquet of two circles. With the identification which is made here, do we have also the same fact ?

I'm assuming that the answer to my preceding question is "yes", in that case, $$\pi_{1}(U_{1})$$ is isomorphic to $$\langle A, B \ | \ \emptyset \rangle$$

Finally, using S-vK theorem, we obtain that $$\pi_{1}(\mathbb{R}\mathbb{P}^{2})$$ is isomorphic to $$\langle A, B \ | \ (AB)^{2} = 1 \rangle$$ and here is my second question. Is this group isomorphic to $$\mathbb{Z} \backslash 2\mathbb{Z}$$ ? It looks like, but I have the feeling that it is not the case.

Also, my last question, we know that the universal cover of $$\mathbb{R}\mathbb{P}^{2}$$ is the sphere $$S^{2}$$. In that case, the group of deck transformations is isomorphic to $$\mathbb{Z} \backslash 2\mathbb{Z}$$. Could you give me explicitely the two elements of this group ?

No! Your $$A,B$$ are not loops (there are no arrows that allow you to identify the head and tail of $$A$$), so they are not generators. The only loop is $$AB$$ (or $$BA$$ if you choose a different basepoint), so your $$\pi_1$$ is generated by $$AB$$, and its square $$(AB)^2$$ in $$\mathbb{RP}^2$$ is killed by your square (sorry, bad pun).

• (+1) I forgot to discuss the issue with their proof, and I think this lays it out perfectly.
– cmk
Jun 7 '19 at 15:43
• Ah ! Yes, I see now... So, if we have only one generator, does it mean that the boundary of the square (with this identification) is homeomorphic to $\langle a \ | \ \emptyset \rangle$ ? Jun 7 '19 at 15:49
• The boundary of the square, $ABAB$ if you read it clockwise starting top-left, is going around the a circle (the $\mathbb{RP}^1$ at infinity) twice. Jun 7 '19 at 15:53

Here's one proof using the lifting correspondence. Let $$p:S^2\rightarrow RP^2$$ be the antipodal identification map. This is a covering map, with $$S^2$$ the universal cover. Let $$\phi:\pi_1 (RP^2,b)\rightarrow p^{-1}(b)$$ be the lifting corresondence. This is bijective because $$S^2$$ is simply connected, and hence $$\pi_1(RP^2,b)$$ only has two elements. Note that this generalizes to $$RP^n$$ for all $$n\geq 2$$.

Another way of using SVK is to remember that $$RP^2$$ can be obtained by gluing a disk the boundary of the Mobius strip. Take $$U$$ to be a neighborhood of the strip and $$V$$ to be a disk. Then, $$U\cap V$$ is an annulus, which has fundamental group $$\mathbb{Z}$$. Since $$U$$ deformation retracts to $$S^1$$, it has fundamental group $$\mathbb{Z}=\langle a\rangle$$, and the fundamental group of $$V$$ is clearly trivial. By SVK, $$f:\mathbb{Z}\rightarrow\pi_1(RP^2)$$ is surjective. We must find the kernel. Let $$k$$ be the inclusion from $$U\cap V$$ into $$U,$$ and $$\ell$$ the inclusion into $$V$$. We only need to consider $$k$$, since the fundamental group of $$V$$ is trivial. If we take the generator $$g$$ for $$U\cap V$$ and send it to the fundamental polygon of the Mobius strip, its image circles twice, meaning that it is homotopic to $$a^2$$. Since the kernel is the smallest normal subgroup containing $$k_*(g)^{-1}\ell_*(g)=(a^2)^{-1},$$ we can conclude that the kernel is the normal subgroup generated by $$\langle a^2\rangle,$$ and so $$\pi_1(RP^2)=\langle a\rangle/\langle a^2\rangle=\mathbb{Z}_2.$$

Making the identifications on just the boundary $$\partial U_1$$ alone does not yield a bouquet of two circles. Also, the four corners of the square are not all identified to each other: the upper-left and lower-right corners are identified to one point, and the upper-right and lower-left corners are identified to another point. Note in particular: neither $$A$$ nor $$B$$ maps to a closed path in the quotient circle, so you may not use either of them as a generator

Instead, the identifications, on $$\partial U_1$$ yield just a single circle which is decomposed into a concatenation of two half-circles $$C=BA$$. We can take $$C$$ as a generator for $$\pi_1$$ of the quotient circle.

Furthermore, the quotient map $$q : \partial U_1 \to S^1$$ is a 2-1 covering map: you can see this by reading the letters in order going clockwise around $$\partial Y_1$$, and you get $$BABA = (BA)^2 = C^2$$.

So now you have enough information to use the same procedure as for the torus and Klein bottle (based on the Seifert-Van Kampen theorem) to obtain the presentation $$\langle C\mid C^2 = 1 \rangle$$ which presents the cyclic group of order $$2$$.