Geometric meaning of r-cycles, r-boundaries and homology groups for a geometric simplicial complex.

Question

I just started learning about algebraic topology, and some things are already not so clear to me. If I consider a geometric simplex $K$, I kind of understand what $H_0(K)$ is. it is a set of equivalent classes [x], where x is a vertex of the simplex, and [x]=[y] means that x and y are in the same connected components. Now for $H_1(K)$ (and more generally for $H_n(K)$, it is not so clear. Elements of $H_1$ are equivalent classes of edges of the simplicial complex, like [xy] or [uv]. [xy]=[uv] means that xy-uv is in the boundary of a 2-simplex. But is geometrically the boundary of a 2-simplex? Also what is a cycle geometrically of a 1,2,3,...-simplex?

Answer 1

Geometrically, think of an n-simplex as an n-sphere ( it makes no difference whatsoever from a topological perspective ). So the n-th homology is simply put "how many ways there are to fit an n-shpere into my topological space". That means, just like the 0-th homology tells you how many parts there are in a space, the n-th lets you know how many n-dimension holes there are in the space you are examining. A cycle is nothing more than an element in the kernel of the boundary operator, homology takes those n-cycles and those n-boundaries and tries to figure out how many of them are redundant ( meaning how many elements in the kernel don't come from higher dimension boundaries ). Now I know that my explanation can be a bit overly simplified, but to understand topology you are heavily reliant in your intuition rather than strict and vigorous mathematical calculations and logical steps due to the nature of the subject. Try to figure things out in a more instrictive rather than a more technical manner.