The real projective plane minus one point is diffeomorphic ... I know that the $2$-dimensional sphere minus one point is diffeomorphic to the plane.
What happens when a point is removed from the real projective plane?
I would like any answer or referencie about It.
Thanks in advanced.
 A: $P = \mathbb{RP}^2$ is obtained as quotient of $S^2$ by identifying antipodal points. Let $p : S^2 \to P$ denote the quotient map. If $y \in P$, then $M_y = P \setminus \{ y \}$ is an open subset of $P$ and hence a non-compact smooth $2$-manifold.
For any two $x,x ' \in S^2$ there exists a rotation $R$ such that $R(x) = x'$. It induces a diffeomorphism $R' : P \to P$ such that $R'([x]) = [x']$. This shows that $M_y, M_{y'}$ are diffeomorphic for all $y,y'\in P$.
We shall show that $M_y$ is a Möbius strip. Let us do it with $y = [1,0,0]$. Then $p^{-1}(y) = \{(1,0,0), (-1,0,0) \}$. The space $S_y = \{(x_1,x_2,x_3) \in S^2 \mid x_3 \ge 0, x_1 \ne \pm 1 \} \subset S^2 \setminus p^{-1}(y)$ is mapped by $p$ surjectively onto $M_y$. The restriction $p' : S_y \to M_y$ of $p$ is a closed map (and hence a quotient map). To see this, let $C \subset S_y$ be closed. Then $C' = C \cup p^{-1}(y)$ is compact, hence $p(C')$ is compact and $p'(C) = p(C) = p(C') \setminus \{ y \}$ is closed in $M_y = P \setminus \{ y \}$. The points identified by $p'$ are pairs of antipodal points in $S_y$ and thus precisely the pairs $(x_1,x_2,0), (-x_1,-x_2,0)$ with $x_1 \ne \pm 1$.
Define
$$h : (-1,1) \times [-1,1] \to D^2 \setminus \{(1,0),(-1,0)\}, h(s,t) = (s,t\sqrt{1-s^2}). $$
It is readily verified that this is a homeomorphism (the segment $\{ s \} \times [-1,1]$ is linearly compressed to the segment $(\{ s \} \times [-1,1])  \cap D^2 = \{ s \} \times [-\sqrt{1-s^2},\sqrt{1-s^2}]$ ). Moreover, the projection $\pi : S_y \to  D^2 \setminus \{(1,0),(-1,0)\}, \pi(x_1,x_2,x_3) = (x_1,x_2)$, is a homeomorphism. Hence $H = \pi^{-1} \circ h : (-1,1) \times [-1,1] \to S_y$ is a homeomorphism and $p'' = p' \circ H : (-1,1) \times [-1,1] \to M_y$ is a quotient map. It is easy to verify that $p''(s,t) = p''(s',t')$ iff either $t = \pm 1$ and  $(s,t) = (-s',-t')$ or $(s,t) = (s',t')$. In other words, $M_y$ is obtained from $(-1,1) \times [-1,1]$ be gluing the boundary components $(-1,1) \times \{-1\}$ and $(-1,1) \times \{1\}$ in opposite direction. The result is a Möbius strip.
