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 $\stackrel{.\,.}{\smile}$

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    $\begingroup$ It's neither. The fundamental group of $\mathbb{R P}^2$ is $\mathbb{Z} / 2 \mathbb{Z}$. $\endgroup$ – Zhen Lin Oct 28 '12 at 21:41
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    $\begingroup$ The fundamental group of $\Bbb R P^2$ is $\Bbb Z/2$... $\endgroup$ – Henry T. Horton Oct 28 '12 at 21:42
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    $\begingroup$ Thanks guys thakt makes abit more sence...:) $\endgroup$ – DifferentialMan Oct 28 '12 at 21:59
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    $\begingroup$ If it makes sense, you should write it up and answer your own question :) $\endgroup$ – Neal Oct 29 '12 at 3:11

You can compute the fundamental group of $\mathbb RP^n$ with aid of the S.V.K. Theorem. Take $\mathbb RP^n$ to be quotient space of $D^n$ so that it's boundary points is identified pairly. So you can use S.V.K. by taking $U=\mathbb RP^n-\mathbb RP^{n-1}$, $V=$ deformation retract neighborhood of it's boundary which deformation retract onto $\mathbb RP^{n-1}$. Then you would conclude that $\pi(\mathbb RP^n)=\pi(\mathbb RP^{n-1})$. For $\mathbb RP^2$, you can give it's $CW$-complex structure to know it's fundamental group. You can also obtain a double-covering from it's universal covering space, i.e., 2-sphere.

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