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 ?
Thank your for your help !
Best Answer
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).