Lack of intuition behind homology theory.

Question

In algebraic topology, we were taught a bit of homology theory such as singular $n$-simplices, singular $n$-chains and homology groups, chain complexes and reduced homology groups. Although we were given some motivation that homology groups identify $n$-dimensional holes, it is not clear to me how exactly it captures the idea.

Why would one think of associating a sequence of abelian groups with a topological space (which I think led to the definition of chain complex), and what does the condition $\partial_{n+1}\circ\partial_n=0$ for all $n\in \mathbb Z$ capture for a chain complex $\{A_n:n \in \mathbb Z\}$ with $\partial_n:A_n\to A_{n-1}$ homomorphisms?

My next question is about the homology groups. What does it mean when we say that homology groups are all those $n$-cycles which are not $n$-boundaries? Is there any reason for calling $Z_n=\text{Ker}\partial_n$ cycles and $B_n=\text{Im}\partial_{n+1}$ boundaries? Why are singular $n$-chains defined to be the formal sum of singular $n$-simplices?

My next question is about chain maps, which are maps between two chain complexes $(A_.,\partial_.^A)$ and $(B_.,\partial_.^B)$ defined by a collection of group homomorphisms $f_n:A_n\to B_n$ where $n\in \mathbb Z$ and the property $f_{n-1}\circ \partial_n^A=\partial_n^B\circ f_n$ is satisfied for all $n\in \mathbb Z$. Also, the concept of chain homotopy between two chain maps is a little complicated.

In short, I am not satisfied with my understanding/visualization of these things. Rather, they seem abstract to me. Can someone provide some help?

Answer 1

I totally get what you're saying. It took me many years to develop some intuition, and I find most (if not all) introductory courses really horrible in that matter.

As often in such situation it is good to look at examples. Consider a triangle. Then its boundary consists of 3 edges. Those edges are arranged in such a way that they form a cycle. This is the fundamental observation behind homology.

So boundaries match with geometric boundaries. Cycles as well. Note that I can put say 5 different edges on the plane. But if they don't form a cycle, a loop, then these are of no interest to us. Because they cannot arise as boundary of triangle(s). Because boundaries are always cycles. This is expressed as $\partial_{n+1}\circ\partial_n=0$ condition.

Now the homology asks the question: if I have 3 edges that form a cycle, then do they arise from a triangle? As you can imagine, this can happen only when there's no hole inside. Well, actually it can be understood as a definition of a hole.

The rest are just technicalities to make it work. In particular groups just fit this setup well, and make all of that easier to compute, rather than actually checking whether continuous extensions exist (which in many cases is extremely hard).

Now as for specific questions:

Why would one think of associating a sequence of abelian groups with a topological space

Because groups (especially abelian) are easier to work with compared to topological spaces.

what does the condition $\partial_{n+1}\circ\partial_n=0$ for all $n\in \mathbb Z$ capture

Boundary of boundary is empty. Or boundary is a cycle. Note that this is not a literal boundary, we take orientation into consideration. So for example if I have two edges: one from point $A$ to $B$, and second from $B$ to $A$, then boundary (of the formal sum of edges) is $A-B+B-A=0$. Here zero represents the "empty" boundary, or rather boundary being cyclic.

Now if we have two edges, say one from $A$ to $B$ and second from $B$ to $C$, with $C\neq A$, then its boundary is $A-B+B-C=A-C$ which is not zero (because $A\neq C$). And so this is not a cycle. And indeed, how can it be a cycle, if it has "ends"? But regardless, if I apply the boundary operator again, I will get zero (note that for $0$-chains it is not that interesting since the entire boundary operator is just zero).

The condition $\partial_{n+1}\circ\partial_n=0$ says that if you arise as a boundary, then you are a cycle. And indeed, boundary of a triangle is a cycle.

What does it mean when we say that homology groups are all those $n$-cycles which are not $n$-boundaries?

As I said earlier, this is the reversed question: if I am a cycle then did I arise as a boundary? Do 3 edges that form a cycle arise as a boundary of a triangle?

Also note that this question is a bit wrong. Homology asks: what $n$-cycles are (or are not) $n$-boundaries? Some of them are, some of them aren't. It may as well happen that all $n$-cycles are $n$-boundaries, which is precisely what $H_n(X)=0$ means.

Is there any reason for calling $Z_n=\text{Ker}\partial_n$ cycles and $B_n=\text{Im}\partial_{n+1}$ boundaries?

I think I answered that earlier as well. The $\partial_n(s)=0$ means that the boundary of $s$ is "empty" (or "trivial"), hence it is a cycle. The $s=\partial_n(x)$ means $s$ arises as a boundary of $x$.

You also should read the definition of $\partial_n$, precisely how it extracts boundary from the simplex and hence why it is called the "boundary operator".

Why are singular $n$-chains defined to be the formal sum of singular $n$-simplices?

Because boundary of a simplex is not a simplex. 3 edges is not a simplex, it is 3 simplexes. We need to express this properly formally. And the idea of formal sums just fits this very well.

My next question is about chain maps

Chain maps is just a way to connect two different chain complexes. Similarly how homomorphisms connect two different groups. Because we would also like to know for example when two chain maps are essentially the same. What does it mean "essentially"? If I relabel elements of a chain complex it is essentially the same, but how do I formally express that?

Also, if we have a continuous map, we would like to know how it "deforms" triangles and affects homologies. It turns out that a continuous map induces a chain map on chain complexes. And this is good, we just convert maps between topological spaces to maps between algebraic objects. We lose some data in the process, but we gain a machinery easier to work with (to compute).

Also, the concept of chain homotopy between two chain maps is a little complicated.

Yeah, it kind of is. A high level of abstraction. I do not know a good intuitive approach to this one. I can only tell you that this condition arises from considering topological homotopies between continuous maps (which induce chain maps). And it is useful even in non-topological context. And it induces isomorphisms on quotient, i.e. homologies. So it is useful.

I hope it helps.