TITLE: How to demonstrate $SO(3)$ is not simply connected?
QUESTION [12 upvotes]: 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?

REPLY [2 votes]: You have two different proofs (a.k.a. the cup proof and the belt proof) right here 
http://www.youtube.com/watch?v=CYBqIRM8GiY
There are several different demonstrations of the second one (Harrison Brown linked to this video in a comment to an answer I submitted - the cup proof - in the question asking for "proofs w/o words").