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I have some questions regarding the real projective plane $mathbbRmathbbP^2$.

If we choose to represent it using the following identification on the square $[0, 1]^2$ :

how can we see that $pi_1(mathbbRmathbbP^2)$ is isomorphic to $mathbbZ backslash 2mathbbZ$ ?

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 $pi_1(mathbbRmathbbP^2)$ is isomorphic to $langle A, B | (AB)^2 = 1 rangle$ and here is my second question. Is this group isomorphic to $mathbbZ backslash 2mathbbZ$ ? 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 $mathbbRmathbbP^2$ is the sphere $S^2$. In that case, the group of deck transformations is isomorphic to $mathbbZ backslash 2mathbbZ$. Could you give me explicitely the two elements of this group ?

Thank your for your help !

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 $mathbbRP^2$ is killed by your square (sorry, bad pun).

Here's one proof using the lifting correspondence. Let $p:S^2rightarrow RP^2$ 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 $ngeq 2$.

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, $Ucap V$ is an annulus, which has fundamental group $mathbbZ$. Since $U$ deformation retracts to $S^1$, it has fundamental group $mathbbZ=langle arangle$, and the fundamental group of $V$ is clearly trivial. By SVK, $f:mathbbZrightarrowpi_1(RP^2)$ is surjective. We must find the kernel. Let $k$ be the inclusion from $Ucap V$ 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 $Ucap V$ 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 $k_*(g)^-1ell_*(g)=(a^2)^-1,$ we can conclude that the kernel is the normal subgroup generated by $langle a^2rangle,$ and so $pi_1(RP^2)=langle arangle/langle a^2rangle=mathbbZ_2.$

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
$$langle Cmid C^2 = 1 rangle
$$
which presents the cyclic group of order $2$.