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The fundamental group of $\Bbb R P^2$ is $\Bbb Z \times \Bbb Z$.

I cannot understand why though, since $\Bbb RP^2$ is a disc with a Möbius strip and the disc is contractible so wouldn't it have fundamental group $\Bbb Z$?

Thanks in advance $\stackrel{.\,.}{\smile}$

  • 5
    It's neither. The fundamental group of $\mathbb{R P}^2$ is $\mathbb{Z} / 2 \mathbb{Z}$.2012-10-28
  • 3
    The fundamental group of $\Bbb R P^2$ is $\Bbb Z/2$...2012-10-28
  • 1
    Thanks guys thakt makes abit more sence...:)2012-10-28
  • 2
    If it makes sense, you should write it up and answer your own question :)2012-10-29

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