Fundamental group of $\mathbb{R}P^2$

Question

I have to compute the fundamentalgroup of the projective plane $\Bbb{R}P^2$ with Van-Kampen Theorem. Therefore I use the fundamental polygon given in Projective Plane. I get for the presentation of the group:

$$\pi_1(X,x_0)=<a,b :abab=1>$$

But this isn't the presentation of $\Bbb{Z}/2\Bbb{Z}$, is it? But this havt to be the result. What is wrong?

I did it this way:

Choose the open set $U$ as the plane without a disc and $V$ as the dics a little bit larger then the discs we ommitted in $U$, thus $U\cap V$ is an annulus. Then I notice $\pi_1(V)=0$ thus trivial and $\pi_1(U\cap V)=\Bbb{Z}$. $U\cap V$ has thus one generator, suppose $g$ (run around the annulus).

An embedding in $V$ gives the empty word (null-homotopical). With an embedding in $U$ is $g$ homotopic to the loop running along the boundary of the polygon. This gives the word $abab$. Thus we get the presentation as above. Where is the mistake? I hope someone can give hints or the good solution. Thanks!

The problem is that $a$ and $b$ are not loops if you take the normal square representation with opposite sides equated by twisting. — Thomas Andrews, Mar 19 '13 at 16:16
Okay but how to formulate the proof if we can not say that $a$ and $b$ are loops? Is $ab$ a loop? — Gerrad13, Mar 19 '13 at 16:18
$ab$ is definitely a loop, because diagonal corners of your square are identified. — Thomas Andrews, Mar 19 '13 at 16:19
The question is, is $\pi_1(V)=\mathbb Z$? That is, is $ab$ the only generator? I don't really know, and my memory of Seifert is very thin. Note that $\leftab:abab=1\right$ is not really a valid representation of a group, because $ab$ is not one symbol. It is really: $\left<r:rr=1\right>$. — Thomas Andrews, Mar 19 '13 at 16:25
I think you can show the $V$ is homotopy-equivalent to $S^1$ pretty easily - $V$ is essentially a mobeus strip, which one can clearly retract to a circle... — Thomas Andrews, Mar 19 '13 at 16:29

Thomas Andrews · Answer 1 · 2013-03-19T16:36:40.743

If you take a square $[0,1]\times[0,1]$ with the identification $(x,0)\sim (1-x,1)$ and $(0,y)\sim (1,1-y)$ then this identification only identifies $(0,0)$ with $(1,1)$, and $(0,1)$ only with $(1,0)$.

So the path $a$ corresponding to $t\to (t,0)$ is not a loop in the real projective plane.

Not sure what the "right" solution to this is for computing $\mathbb RP^2$.

If you take $U$ to be the image of $(0,1)\times (0,1)$ and $V$ to be the image of $T^2$ where $T=[0,\frac{1}{3})\cup(\frac{2}{3},1]$, $U$ is clearly simply connected and I think it is easy to show that $V$ is retractable to a circle - the image of $T^2$ is a Möbius strip.

These are essentially the same $U,V$ you chose, but I think picking a square $U$ makes it more obvious what $V$ is.

@Gerrad13 If you found this answer helpful, please accept it. — A.P., Mar 19 '13 at 18:02

Fundamental group of $\mathbb{R}P^2$

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