How do homology groups work? Looking at the wikipedia article, it lists, for example, $H_k(S^1) = \mathbb Z$ for $k = 0,1$ and ${0}$ otherwise. It also says that $H_k(X)$ is the k-dimensional holes in $X$. Thus, there is a 0-d hole and a 1-d hole. I see a 2-d hole, but neither of 1-d nor 0-d. This trend continues for the other example listed. What is it that I am completely missing? Thank you.
[Math] How do Homology Groups work
algebraic-topologyhomology-cohomology
Best Answer
Actually, homology groups of a k-dimensional space do measure holes, but for dimensions $1,2,..,k-1$. For dimension zero, they measure the number of connected components , and for the top homology, they measure whether the space is orientable or not. A hole is informally defined as an obstruction to shrinking an n-dimensional object within the space, so that a 1-d hole in a topological space of dimension larger than one an obstruction to shrinking a curve within the space into a point, though ( see comments) you may consider a non-zero top -dimensional homology as a hole .
Notice that, algebraically, we define a hole to be a cycle that does not bound, i.e., we say that the homology is non-trivial , or that there is an n-hole if the quotient $Z_n/B_n \neq {id}$. If you look, e.g., at the case of a 2-torus $T^2:= S^1 \times S^1$ , you will see that, e.g., a meridian is a cycle that does not bound, because its removal will not disconnect the space. Similarly for any strictly latitudinal curve. These two cycles (simple-closed curves in the space) generate the homology of the torus.
An additional property of homology groups is that EDIT: this applies to most spaces you will run into unless you do specialized work ( see Mariano's comment below) , for a space $X$ of topological dimension: http://en.wikipedia.org/wiki/Topological_dimension greater than $k$, we have that $H_k(X) =0$. Notice this is not true for some of the groups $\pi_n(X)$.