# Why Are Homology Groups Not MUCH Larger?

1. May 16, 2014

### Mandelbroth

So far, I think algebraic topology is turning out to be the best thing since sliced bread. However, I'm having a bit of difficulty with homology, for one particular reason.

Consider, as an example, the first homology group of $S^1$. The definition of the free abelian group (or, in general, the $R$-module) of $n$-chains is the free abelian group generated by ALL $n$-simplices. Why do we not include ALL paths in $S^1$ in this definition? Are these not also 1-simplices?

Thank you!

2. May 16, 2014

### HallsofIvy

You are not talking about "paths", you are talking about equivalence classes of paths. Many different paths lie in one equivalence class.

3. May 16, 2014

### Mandelbroth

AHA! Thank you! That makes soooooo much more sense. :rofl:

4. May 16, 2014

### WWGD

To add a little bit, remember that homology is a quotient be it of modules or vector spaces --cycles/boundaries -- which shows the elements to be classes, where two cycles are equivalent (geometrically *) if their difference bounds . And in the case of $S^1$ , the classes are the paths that go around n times. You can see that if you go around a non-integer "number of times" k , then the path can be deformed/homotoped to the path class |_ k _|, where |_ k _| is the greatest integer less than k .

* You can also do this purely algebraically.