# Sweating out the homology

This is going to be a rather long post on homology. I hope I do manage to understand it. It will ultimately go up in a polished form on my blog. The reason why it is difficult to understand homology and cohomology without typing it all out is that the information given is so little. One has to construe so much from relatively dry language. I think that is the place where writing things out would help tremendously.

If two spaces $X$ and $Y$ are homotopic, then their homology groups are isomorphic. If the homology groups differ, then they clearly cannot be homotopic, and more specifically, homeomorphic.

A chain map $f:C^*\to D^*$ is a map between homotopy groups $C_n$ for each $n$. Hence, a chain map encodes information about an infinite number of maps between those homotopy groups. Also, these maps commute with boundary maps. What does it really mean for a map to commute? It means that the maps $\partial$ and $f_n$ literally commute! $\partial f_n=f_n\partial$ for all $x\in C_n$. Hence the name commutative diagrams. This is honestly the first time that I have thought of this, in spite of having read about commutative diagrams all this bloody while. Maybe typing does have its benefits.

What kinds of maps take cycles to cycles and boundaries to boundaries? Commuting maps definitely do. But the commuting structure has to be present throughout. Above and below. Do other kinds of maps have similar properties? Yes. It is possible. In such a commuting structure, we have a natural map between homology groups. It is only because of the commutativity that the map is well defined; i.e. that elements in $B_n$ go to $C_n$.

We now construct a functor: for a chain map $f: C_*\to D_*$, we construct a functor $H_n$ such that $H_n (f): H_n (C_*)\to H_n (D_*)$. As far as the mappings of objects go, we see a mapping of $C_n$‘s to $H_n$‘s. Do we have to have an infinite number of functors to be able to successfully create all maps between the images of chains? Yes. We need a separate functor for each map $f: C_n\to D_n$. $H_n$ is known as the $n$th homology functor.

Now we talk about chain homotopic maps. Two maps $f,g: C_*\to D_*$ are chain homotopic if there exist maps $h_n: C_n\to D_{n+1}$ such that $f_n-g_n=h_{n-1}d_n+d_{n+1}h_n$. What does it mean for two maps to be chain homotopic? And why should such $h_n$‘s exist? In understanding this, this answer came in handy. Essentially, two chain homotopic maps induce the same maps between homological groups. Hence, homotopic chain maps induce isomorphic homological maps. Why is that? This is because $f(z)-g(z)$ for $z\in$ Ker $\partial _n$ belongs to $B_n (D_*)$. Hence, $f$ and $g$ induce the same maps between the homology groups.

Just a quick note: the kernels and the cokernels in the snake lemma and other lemmas are all with respect to the boundary maps, and not the chain maps. Why is that? Because the chain maps just act like connecting and commuting linkages. The main action happens with sets related to the boundary maps- like the kernels and the cokernels.

Now I shall study the Snake Lemma. With this lemma, there is one thing that has always confused me- how is the map well-defined? What if their difference goes to $0$? What does all this mean? OK. First of all, in whichever direction we go in the process of diagram chasing, we might or might not have a well-defined choice. We always have to check for a well-defined choice. Checking in some instances is easier than checking in others. In this case, everything works out as the lower left map is injective, and the whole diagram is commutative. Sorry this is not a completely rigorous or complete explanation, but it has made me understand something that I was at a loss to understand for far too long.

Now we shall talk about the long exact homology sequence. Say you have a short exact sequence of chain complexes $0\to C_*\to D_*\to E_*\to 0$. How is the long exact homology chain induced? The main issue here is the construction of the connecting map $H_n(E_*)\to H_{n-1}(C_*)$. This is done by using Snake’s lemma on the following diagram 