TITLE: Fundamental group of real projective plane minus one point
QUESTION [8 upvotes]: I understand that the $\mathbb{R}P^2$ is homeomorphic to the unit disc with boundary points identified with their antipodes. But even if we puncture the disc and stretch it from the origin to let it be retracted to the boundary, how are we gonna justify that the identified boundary is still homeomorphic to $S^1$(or is it?)?

REPLY [5 votes]: The punctured $\Bbb{RP}^2$ deformation retracts onto the boundary circle of the disc with antipodal points identified, $\Bbb{RP}^1 = S^1/(x \sim -x)$. This is homeomorphic to the circle, by the map $S^1 \to \Bbb{RP}^1$, $z \mapsto z^2$. This is a continuous bijection etc so a homeomorphism, as desired, and thus the punctured projective plane is homotopy equivalent to $S^1$, and has fundamental group $\Bbb Z$.
(One can be far more precise than this: the punctured plane is homeomorphic to the (open) Mobius band.)