>>10146328

>S^1 being homotopic to Z

S^1 is not homotopic to Z, it's fundamental group is Z and its homology in degree 0 and 1 is Z, which is something completely different to that.

Homology groups in degree n tell you about how many n dimensional holes there are in your space, and how they behave.

When looking at homology with integer coefficients (what you're doing), the 0th homology tells you about the amount of path components, that is, how many distinct points are there that you can't draw a path between them. The 1st homology tells you about 1 dimensional holes, and S^1 has one "hole". The homology is Z because of the way you build chain complexes. If you were to have homology with coefficients in an arbitrary abelian group G, you'd get that the first homology of S^1 would be G.

You shouldn't think of Z in the same sense as the fundamental group. Instead, look at the exponent of Z. If you had a space with [math]H_0(X)=\mathbb Z^3[/math] and [math]H_1(X)=\mathbb Z^2[/math], then you can exactly say that your space has 3 path components, and roughly two 1 dimensional holes. An example of a space that gives this is if you had an isolated point, and two isolated circles, or two isolated points and a space that looks like an 8. A final example is an isolated solid disk, an isolated point, and a torus (the latter has one hole in the crosssection and another going around it).

But homology doesn't just capture holes, it also captures "twist" or torsion. For example, projective 2-space is defined as the sphere with opposite points identified. Hence, when you look at the equator of this sphere, going around halfway around the equator is actually the same as going around the equator once, so going once around the whole equator is the same as going around the equator twice. This "degree 2" idea is captured in the fact that [math]H_1(\mathbb RP^2)=\mathbb Z/ 2\mathbb Z[/math], and similarly if you had "degree n" maps you'd get [math]\mathbb Z/n\mathbb Z[/math] terms.