# Projective plane - {0} homeomorphic to cyllinder?

1. Apr 16, 2010

### wasia

1. The problem statement, all variables and given/known data
Assuming that $$\mathbb S^1 \times (0,1)$$ is not homeomorphic to $$\mathbb R^2$$ show that the sphere $$\mathbb S^2$$ is not homeomorphic to the projective plane $$\mathbb P^2$$.

2. Relevant equations
N/A

3. The attempt at a solution
This asks for a proof by contradiction - assuming $$\mathbb S^2 \cong \mathbb P^2$$ ("$$\cong$$" meaning "homeomorphic"), we remove a point from $$\mathbb S^2$$ to get $$\mathbb R^2$$. However, I cannot prove that by removing a point from the projective plane we get a cyllinder. I have considered $$\mathbb P^2$$ both as $$\mathbb S^2 / \mathbb Z_2$$ and producing it from the $$\mathbb R^3$$. Any help appreciated!