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I am reading the book Application of Path integrals by Schulman, which has a chapter on applications of homotopy theory to path integrals. In that he says we can geometrically describe $SO(3)$ by a solid (3 -dimensional) ball with radius $\pi$,and with antipodal points identified. Each point in the ball at distance $\phi$ from the centre, represents a rotation about the axis passing through that point and origin, and angle of rotation $\phi$. Later he writes "Projective 3 space is homeomorphic to the solid ball described earlier." I imagine $\mathbb{RP}^3$ to be the 3-sphere with antipodal points identified.

How is this homeomorphic to the solid ball described above?

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It might help the intuition by considering the case of $\mathbb{RP}^2$ first: it can be seen either as the quotient or identification space of $S^2$ where antipodal points are identified, or as the quotient space of a solid 2-disk $D^2$ with antipodal points on the $S^1$ boundary identified. We are just making an analogous statement in three dimensions.

My own preferred definition of projective space is the set of lines through the origin (topologized appropriately). If we consider first a sphere centered at the origin, then each such line intersects the sphere in two antipodal points, and these are thus identified as in the first model. On the other hand, we could consider just the northern hemisphere (the set of points on the sphere where $z \geq 0$) and play the same game; here each line hits the hemisphere just once, unless the line lies in the plane $z = 0$, where it hits the equator twice in two antipodal points, which are thereby identified. This gives the second model (since the hemisphere is homeomorphic to a solid $2$-disk $D^2$).

The same carries over in three dimensions. I doubt you need the formal details, although they could be supplied if need be.

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