Let \(X\) be the quotient space obtained from \(B^2\) by identifying each point \(x\) of \(S^1\) with its antipode \(-x\). Show that \(X\) is homeomorphic to the projective plane \(P^2\).
Solution
Recall that the real projective plane is \[ P^2=S^2/(x\sim -x). \] Let \[ S^2_+=\{(x_1,x_2,x_3)\in S^2:x_3\ge0\} \] be the closed upper hemisphere. Projection onto the first two coordinates gives a homeomorphism \[ \phi:S^2_+\longrightarrow B^2, \qquad \phi(x_1,x_2,x_3)=(x_1,x_2), \] with inverse \[ (u,v)\longmapsto (u,v,\sqrt{1-u^2-v^2}). \] Every antipodal pair in \(S^2\) has exactly one representative in the open upper hemisphere; pairs lying on the equator have two representatives in \(S^2_+\), and these are antipodal equatorial points. Under \(\phi\), the equator corresponds to \(S^1=\partial B^2\), and the antipodal relation on the equator becomes exactly \[ x\sim -x\qquad(x\in S^1). \] Thus the quotient map \(S^2\to P^2\), restricted to the upper hemisphere, identifies precisely the same boundary pairs as the quotient \[ B^2/(x\sim -x\text{ on }S^1). \] Therefore it induces a continuous bijection \[ B^2/(x\sim -x)\longrightarrow P^2. \] The domain is compact and \(P^2\) is Hausdorff, so this bijection is a homeomorphism. Hence the quotient space in the exercise is \(P^2\).