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 $[0, 1]^{2}$ :

enter image description here

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 ?

Thank your for your help !


3 Answers 3


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).

  • $\begingroup$ (+1) I forgot to discuss the issue with their proof, and I think this lays it out perfectly. $\endgroup$
    – cmk
    Jun 7, 2019 at 15:43
  • $\begingroup$ 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 $\langle a \ | \ \emptyset \rangle$ ? $\endgroup$ Jun 7, 2019 at 15:49
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 7, 2019 at 15:53

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 $2$.


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 $\pi_1(RP^2)=\langle a\rangle/\langle a^2\rangle=\mathbb{Z}_2.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.