I'm trying to compute the fundamental group of $\mathbb{RP}^2$ minus 2 points. I'm using the presentation $\langle a\mid a^2\rangle$. Meaning that I'm taking the disk and identifying the two sides.

I'm conjecturing that it is $\mathbb{Z}\times\mathbb{Z}$. I'm trying to use Seifert -Van Kampen but I can't get my open sets to work nicely. The reason why I think it is $\mathbb{Z}\times\mathbb{Z}$ is because clearly a loop around each hole would give you two distinct loops, say $a$ and $b$. But then it seems that $ab$ is homotopic to $ba$. This relation leads me to think that it might be $\mathbb{Z}\times\mathbb{Z}$.

Any hint would be appreciated.

  • $\begingroup$ The fundamental group is $\mathbb{Z}$, not $\mathbb{Z}\times \mathbb{Z}$. $\endgroup$ – Bombyx mori Jan 7 '15 at 0:32
  • 3
    $\begingroup$ The projective plane minus a point/small disc is homeomorphic to a Mobius band. Show this, and proceed from there. $\endgroup$ – user98602 Jan 7 '15 at 0:33
  • $\begingroup$ @Bombyxmori: OP said minus two points, so is a Möbius band without a point $\endgroup$ – janmarqz Jan 7 '15 at 2:52

Here is the standard way of tackling such problems:


Think of $\mathbb{RP}^2$ as as the disc $D^2$, with boundary identifiead in the oposite direction. Then after removing $2$ points the space retracts onto the space $X$. Now you can easily see that $X$ is nothing but $S^1\vee S^1$. So you have, $\pi_1(\mathbb{RP}^2\setminus\{\text{two points}\})\cong\pi_1(X)\cong\mathbb{Z}*\mathbb{Z}$

  • $\begingroup$ The picture makes so much sense. $\endgroup$ – Phanu9000 Jan 7 '15 at 16:20
  • $\begingroup$ Would be able to elaborate on this retraction? $\endgroup$ – user475040 Jan 9 at 21:37
  • $\begingroup$ @crispypizza The (deformation) retraction is simply "radially" pushing everything away from the two puncture points. You'll end up with the space $X$. If you're asking for an explicit map, that would be quite messy! $\endgroup$ – ChesterX Jan 10 at 5:06
  • $\begingroup$ Yes definitely not, I guess I am not sure where the segment connecting $u$ is coming from. We are getting another generator $b$? $\endgroup$ – user475040 Jan 10 at 20:56
  • $\begingroup$ @crispypizza It is one of the generators. If you consider the horizontal segment before performing the retraction, it is a non trivial loop. $\endgroup$ – ChesterX Jan 11 at 6:23

This is akin to the approach taken by Bombyx mori. Let $\mathbb{RP}^2$ be identified with $S^2$ under the quotient of antipodal points. Thus we are removing two pairs of antipodal points and then quotienting. Let those points be $$(\sqrt{2}/2, \sqrt{2}/2,0), (-\sqrt{2}/2, \sqrt{2}/2,0), (\sqrt{2}/2, -\sqrt{2}/2,0), (-\sqrt{2}/2,-\sqrt{2}/2,0)$$ i.e. evenly spaced around the equator. Then there is a deformation retract of $S^2 - \{ \text{points}\}$ onto the union of great circles $$A = \{(x,y,z) \in S^2 | x = 0\} \cup \{ (x,y,z) \in S^2 | y = 0\}$$ Furthermore, we can arrange it so this deformation respects the antipodal quotient. But this descends then to a deformation retract $$ \mathbb{RP}^2 \to A/\{\pm\} \cong S^1 \vee S^1$$ And this has fundamental group $\mathbb{Z} * \mathbb{Z}$.


Here might be an unconventional way to see this to get around the Mobius band. Let $\mathbb{RP}^{2}$ to be $\mathbb{S}^{2}$ with antipodal points identified. Now removing four points, with two pair of them anti-podal then take the quotient is the same as removing two points from $\mathbb{RP}^{2}$. But the former one is easy to describe - it is $\mathbb{S}^{1}\vee \mathbb{S}^{1}\vee \mathbb{S}^{1}$. Now by taking the quotient we identify two such circles into one circle. So in the end we get $\mathbb{S}^{1}\vee \mathbb{S}^{1}$, and the fundamental group is $\mathbb{Z}*\mathbb{Z}$.

In this light my earlier comment (as janmarqz pointed out) was wrong, as the comments showed I could not see this clearly until thinking more carefully about it. In fact the Mobius band picture is equally clear, and I was confused with the extra $\mathbb{S}^{1}$ coming from the shrinking part of the Mobius band.

  • $\begingroup$ excuse me, man, if I was too rude, but you already see, (+1) for all $\endgroup$ – janmarqz Jan 7 '15 at 4:27
  • $\begingroup$ @janmarqz: Yes, by the point you posted your comment I knew it was $\mathbb{Z}*\mathbb{Z}$ already. So I want some reasoning from others to convince me I was false by assuming it should be a circle. Since I got zero feedback I have to think more carefully. It is okay. $\endgroup$ – Bombyx mori Jan 7 '15 at 4:30
  • $\begingroup$ @janmarqz: Forgot to say that thanks for pointing out. $\endgroup$ – Bombyx mori Jan 7 '15 at 4:31
  • $\begingroup$ you're welcome. $\endgroup$ – janmarqz Jan 7 '15 at 4:33

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