Sphere bundle of the tangent bundle of 2-dim sphere Let sph($\tau S^2$) be the sphere bundle of the tangent bundle of 2-dim$^l$ sphere. Could someone tell me why  
sph($\tau S^2$)=$\mathbb{R}$P$^2$$\cup$$e^3$
holds? Where $e^3$ is a 3-dim$^l$ cell.
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$Here's a sketch:


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*The unit sphere bundle of $TS^{2}$ is homeomorphic to the rotation group $SO(3)$. (Idea of proof: An element of $SO(3)$ is essentially an ordered orthonormal basis of $\Reals^{3}$; projection to the first column is a map to $S^{2}$; the second column gives a unit vector in the tangent space at the first column. Because $SO(3)$ is a compact Hausdorff space, the preceding continuous bijection has continuous inverse.)

*The rotation group is homeomorphic to the three-dimensional real projective space $\Reals\Proj^{3}$. (Take a closed ball of radius $\pi$ in $\Reals^{3}$; map the center to the identity; map each non-zero point $x = |x|\cdot(x/|x|)$ to counterclockwise rotation by angle $x$ about the unit vector $x/|x|$. Show this map is continuous, bijective on the open ball, and identifies antipodal boundary points. As in the preceding bullet point, the map is consequently a homeomorphism between $SO(3)$ and the closed ball with antipodal boundary points identified.)

*The result of attaching a three-cell to the real projective plane $\Reals\Proj^{2}$ (via the antipodal map on the boundary sphere) is $\Reals\Proj^{3}$. (See preceding point.)
