Map R onto S, the points of the sphere, by considering only the relocation of the south pole. This can be determined by running a unit vector, i.e. the axis that corresponds to the south pole, through the matrix that implements the rotation. This is a quotient space, and the map is bicontinuous. The fiber is the rotations that fix the south pole, like a planet spinning on its axis. This is homeomorphic to a circle.
The map from R onto S is a fiber bundle. Each neighborhood of R looks like a region of S, where the south pole might wander, crossed with a circle, which determines how the sphere spins as its pole is fixed within that neighborhood.
All this works in higher dimensions. The rotations of Sn map onto Sn, with a fiber of the rotations of Sn-1.
In higher dimensions, the path is once again pushed down to the south pole, where it becomes a path through the rotations of the n-1 sphere. The fundamental group is thus generated by the fundamental group one dimension down. So perhaps we can bootstrap up to higher dimensions; but first we must begin with the fundamental group of the rotations of the 2 sphere, which is generated by homotopic paths through the rotations of the 1 sphere, and those paths are the integer windings around the circle.
The long exact sequence confirms what we already know. π1(S1) → π1(R) → π1(S2), and the last group is 0. When a loop spins around the circle d times, it becomes an arrow that spins around the south pole d times, and this covers all the loop classes of R. The fundamental group of R is a quotient group of Z.
I'm going to present an intuitive argument for Z. This is not a proof, but it's a good start.
Assume, at each time t, the homotopy presents a smooth loop in S. We never run into one of those fuzzy space-filling curves. The loop has a direction as you move along in S. The rotational arrow indicates another direction. Let θ be the difference between these two directions. This is a continuous function along the path, and it starts and ends at the same value, i.e. the direction of the arrow at the south pole when the sphere is fixed, minus the direction of the loop as it leaves and reenters the south pole. The change in θ, as you move around the loop, is d times 360 degrees. Here d is the degree of the loop. Throughout the homotopy, arrows change direction continuously, and the loop changes direction continuously. This means θ changes continuously. A homotopy on θ cannot change the degree, hence d is constant from start to finish. You can't spin once around the south pole at t = 0, and 17 times around the south pole at t = 1.
Of course we have a problem at t = 0, because the path has no direction. For each length l along the path, artificially assign it the direction that it has, at position l, as t approaches 0. Do the same for t = 1.
If a loop at time t has corners, smooth them out. If a loop stops for a while, reparameterize. Thus any normal homotopy, the kind that looks like a loop twisting and sliding around the sphere, will not change the degree of the spinning arrow at the south pole. And yet, I'm going to build just such a homotopy. We'll figure out where the degree argument breaks down later.
Here is a homotopy from the arrow spinning around the north pole twice, to the arrow fixed at the south pole. After that you can always slide the arrow up the front of the sphere and back to the north pole to complete the homotopy. I will present the equations first, then try to explain them in words.
The homotopy proceeds from t = 0 to 1. The variable r is shorthand for sqrt(1-t2). The loop is represented by θ, running from 0 to 2π. The columns of the following matrix are x y and z. Picture z as pointing up along the north pole. The x axis runs to the right, and the y axis runs away from you, as you are facing the front of the sphere. In the following matrix, the bottom row is the image of the north pole, and the top row is the image of a vector that was originally perpendicular to the north pole. In other words, the bottom row is the new location of the north pole on the sphere, and the top row is the direction of the arrow.
| cos(2θ)r2 + t2 | sin(2θ)r2 | 2sin(θ)rt |
| -sin(2θ)r2 | cos(2θ)r2 - t2 | 2cos(θ)rt |
| 2sin(θ)rt | -2cos(θ)rt | r2 - t2 |
Each of the 9 entries is continuous in t and θ. So we have a continuous map from the circle cross time into 3 space.
Brush up on your trig, and prove each row has length (r2+t2)2, which is 1. Then prove the dot product of every pair of rows is 0. The matrix is orthonormal. You might still be worried about stumbling upon a reflection, rather than a rotation, as indicated by a determinant of -1, rather than 1. Expand the determinant and simplify, giving (r2+t2)3, which is 1. The image is a rigid rotation at all times, and the entire homotopy lies in R.
When t = 0, the bottom row is the vector [0,0,1], i.e. the north pole. The top row spins around twice as θ moves around the circle once. This is the start of the homotopy.
When t = 1, the bottom row is [0,0,-1], i.e. the south pole. For all values of θ, the top row is [1,0,0], i.e. an arrow pointing right. That completes the homotopy - but what happened to my degree argument? Let's try to understand the homotopy at an intuitive level.
Stop the homotopy at time t, and the bottom row tells you what the loop looks like, without regard to the arrows. The z coordinate is fixed, and the x and y coordinates trace out a horizontal circle that starts and ends at the front of the sphere. In other words, the loop is a circle of lattitude, and as t moves from 0 to 1, the circle expands outward from the north pole, sweeps down across the entire sphere, and closes in on the south pole. When t = sqrt(½) the circle is the equator.
When the loop first leaves the north pole, t is nearly 0, and r is nearly 1. The loop travels around the north pole once, and the arrow spins around twice. Relative to the direction of travel, the arrow spins around once, and the degree is 1.
As the loop approaches the south pole, t is nearly 1, and r is nearly 0. The arrow points right at all times, with some small wiggles here and there. Move around the circle, and the arrow turns relative to the direction of travel. In fact it turns once around counterclockwise. The degree is still 1. By sliding over the sphere, the path has, in some sense, reversed its direction. Near the south pole the spinning arrow cancels the path's rotation, instead of adding to it. It's beautiful, isn't it!
Just for grins let's see what hapens at the equator, when r = t = sqrt(½). The arrow vector is half of [cos(2θ)+1,sin(2θ),2sin(θ)]. This starts out pointing right, and tips up to vertical as you move a quarter of the way around the circle. Half way around and it is pointing right again. After the third quarter it points down. And of course back to pointing right at the start. Watch the arrow as you travel around the circle, and the degree is still 1.
Step up to the second row of our long exact sequence, which beginse with π2(S1), or 0. Connect the next two groups with the one below to build the following short exact sequence.
0 → π2(R) → π2(S2) → Z → 0
The last group is represented by the loops in the circle with even degree. These lie in the kernel of the next homomorphism, hence they must be the image of the previous homomorphism. A homotopy class of the sphere in the sphere pulls back to infinitely many homotopy classes of the sphere in R.