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# How to demonstrate $SO(3)$

A quote from Wikipedia's article on the Rotation group:

Consider the solid ball in $\mathbb{R}^3$ of radius $\pi$ [...]. Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball. Rotation through angles between $0$ and $-\pi$ correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through $\pi$ and through $-\pi$ are the same. So we identify [...] antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group.

So far, so good. This illustrates $SO(3)\cong \mathbb{RP}^3$.

These identifications illustrate that $SO(3)$ is connected but not simply connected. As to the latter, in the ball with antipodal surface points identified, consider the path running from the "north pole" straight through the center down to the south pole. This is a closed loop, since the north pole and the south pole are identified. This loop cannot be shrunk to a point, since no matter how you deform the loop, the start and end point have to remain antipodal, or else the loop will "break open".

I believe that $SO(3)$ is connected but the "intuitive argument" for $\pi_1(SO(3))\neq 0$ is not clear to me: The starting point at the "north pole" is a rotation of $\pi$ counterclockwise around the $z$ axis. This agrees with the "south pole", a rotation of $\pi$ clockwise around the $z$ axis. So the described path is a full $2\pi$ rotation counterclockwise around the $z$ axis, stating not in the identity position. Why isn't this homotopic to the trivial path? Antipodal points are identified, so what does "start and end point have to remain antipodal, or else the loop will "break open"" mean?