Why Are Homology Groups Not MUCH Larger?

[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!

 

[HallsofIvy]

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

 

[Mandelbroth]

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

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

 

[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.