To get us started, consider one simplex, rather than a collection of simplexes. A simplex is a cycle with 0 boundary iff it is a closed path, or a loop. So you might think h1 is related to π1, the fundamental group of S. They are related, but not identical. As it turns out, h1 is based on the conjugacy classes of π1. Recall that these classes correspond to the homotopy classes of the circle in S. Homotopic loops are homologous (shown below), so all the homotopic loops combine to form the same entity in h1.
Consider the standard triangle. The boundary of the standard 2-simplex consists of a positive path along the x axis, another positive path flowing up the hypotenuse, and a negative path going up the y axis. Picture the positive paths as red arrows and the negative path as a blue arrow. If a 2-dimensional simplex in S has corners x y and z, the boundary is a red arrow (could be a squiggly path) from x to y, a red path from y to z, and a blue path from x to z. We're going to use these triangular simplexes to turn homotopy into homology.
If you are interested in homology, there is a way to get rid of blue paths. Suppose there is a blue path from x to z. When viewed as a linear combination of 1 simnplexes, this path has a coefficient of -1, hence the color blue. Create a new 2 simplex that is squashed onto the path from x to z. The vertices map to x, z, and x respectively. The boundary is a red path from x to z, a red path from z to x, and a blue path from x to x. The latter is the trivial map that carries an interval onto x. Since this is a boundary, we can add it to our original picture without changing the homology. Red and blue paths cancel, leaving a red path from z to x and a blue dot at x. Thus any blue path can be reversed and changed to red, if you don't mind a blue dot popping into existance. These dots can go away, because any point is the boundary of the standard triangle mapped onto that point. Therefore we can assume all paths in a chain have positive coefficients.
Assume l1 and l2 are loops in S that are homotopic. Let f be the homotopy, a continuous map from S1 cross I into S with l1 at the bottom and l2 at the top. Let c be the base point of S1 at the bottom of the cylinder, and let d be the corresponding base point at the top. Draw a segment from c up to d, and a spiral from c, all the way around the cylinder, and up to d. This cuts the wall of the cylinder into two triangular simplexes: a = cdd (dd going around the top) and b = ccd (cc going around the bottom). Consider the boundaries of a and b. The counterclockwise paths, around the top and bottom of the cylinder, are red. The segment from c to d is red, and the spiral from c around to d is blue. Apply f to the difference a-b, and the boundary is l2-l1. Both loops are in the same coset of bound(G2), and represent the same entity in h1.
Don't assume the converse. Loops that are not homotopic may nonetheless be homologous. They may form the boundary of a collection of 2 dimensional simplexes. We'll see this later on.
Of course a cycle could comprise more than one simplex. When the simplexes are paths, a cycle is a digraph, running through S, such that the in-degree of each vertex equals its out-degree. Assume all the coefficients on the paths are positive. (All the paths are red.) Start at a vertex and follow the paths, until a vertex repeats. This is a cycle, in the graph theory sense. It is also a cycle with respect to homology. You pass through each vertex once, and the boundary contributes +1 from the incoming path and -1 from the outgoing path. The boundary is 0, as expected. Do this again and again, until the digraph is a sum of such cycles. Thus these cycles are sufficient to generate the digraph. A cycle might be a loop, or a succession of paths that form a loop, such that each vertex is passed through only once.
These multipath versions of a loop are not necessary, as each is homotopic to a loop. The homotopy expands the first arc all the way around the loop, squashing the other arcs down to a point in the process. Return to our can, which implements the homotopy. This time the top and bottom are both l1, but the top is cut into several arcs. Draw vertical segments down the wall of the can, according to the vertices that delimit the arcs on top, then push these lines around the can, so they become spirals that terminate at the base point c. The first arc, at the bottom of the can, has effectively pushed its way around the circle, pulling the segments along with it. Each arc at the top of the can is now part of a triangle, except for the first arc, which is part of a quadrilateral. Draw in one more diagonal, from c up to the end of the first arc, and we're there. Let arrows point down the sides of each triangle, blue on the left and red on the right. Clearly the sides of the triangles will cancel. The last triangle, at the bottom of our quadrilateral, is red on the left and blue on the right. Subtract a triangle that is red on the left and blue on the right and red around the bottom loop. That does the trick. Thus there is no need for the multipath loops. One path - one instance of l1; that is enough to generate h1.
In summary, h1 is generated by the homotopically distinct loops in S, which are the conjugacy classes of π1(S).
The fundamental group is Z, which is abelian, hence the conjugacy classes are the members of the fundamental group, or Z itself. Let the standard map, zd in the complex plane, wrap the circle around itself d times. This represents the class of loops having degree d. These loops generate h1.
Returning to our standard triangle, let each red arrow wrap around the circle once, and let the blue arrow wrap around the circle twice. Both paths wind around the circle twice, and are homotopic. Thus the entire triangle can be mapped onto the circle, consistent with these boundary conditions. Take the boundary, and z2 = 2z. In other words, z2 is not an independent generator after all. It is twice the generator z.
Let the red leg = z, let the red hypotenuse = z2, and let the blue leg = z3. Again, a homotopy allows us to fill in the triangle. This proves z3 = z+z2 = 3z. The generator z3 is not free either. This extends, by induction, to every degree. Similar reasoning shows z-d + zd = 0, hence the negative degree loops are not free either. Therefore h1 is generated by a single loop, namely z, which is the identity map on the circle.
Suppose a multiple of z, such as 17z, is the boundary of a linear combination of 2 dimensional simplexes. Remember that all coefficients can be changed to 1. Ten times a simplex is the same as ten times around that simplex, and -1 times a simplex is a path around that simplex going the other way. So we just have a sum of triangular paths. Some of the edges are z, net 17 to be precise, and any other edges must cancel. Take one of these triangles and let it define a homotopy, thus replacing one edge with the other two. Use this technique to kill off one of the 17 trips around z, and you still have a connected loop, homotopic to 17z. Any new edge must be canceled, and we still have 16 instances of z to kill off. Apply the next triangle, and the next, until the last edge is gone. The succession of finitely many homotopies, one per triangle, builds a larger homotopy that pulls 17z down to 0. A loop generates a subgroup of order d in h1 iff d times around that loop is homotopic to a point. With this in mind, we only need consider the distinct conjugacy classes of a loop and all its multiples in π1(S). And this applies to any space, not just the circle. But when S is the circle, each zd is homotopically distinct; therefore h1(S1) = Z, the free abelian group whose generator is the identity map on the circle. If you prefer, the reduced homology is 1.
All higher spheres are simply connected, and S0 is two simply connected points. Thus h0 and h1 are known for all spheres. Reduced h0 is 2 for S0, and 1 for all higher spheres. Reduced h1 is 1 for S1, and 0 for all other spheres. Subsequent pages will derive all the homology groups for all the spheres. It's going to be fun.