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A Homology Theory is a Monster
Bredon's Plan of Attack: State all, apply all, prove all.
Our Route: Axiom by axiom - state, apply, prove. Thus everything we will do will be, or should be, labeled either "State" or "Prove" or "Apply".
The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.
Recall we had defined for a chain complex the associated homology groups:
From this we get the pth homology for a topological space
We have previously shown that
1) for disjoint unions of spaces
4) via the map
where is a path connecting to x.
We need to check the maps are in fact inverses of each other.
Lets consider . We start with a closed path starting at (thought of as in the fundamental group). means we now think of it as a simplex in X with a point at . now takes this to the path that parks at for a third of the time, goes around the loop and then parks for the remaining third of the time. Clearly this is homotopic this composition is homotopic to the identity.
We now consider . Start with just a path . Then makes a loop adding two paths from the to the start and finish of forming a triangular like closed loop. We think of this loop as
Now, we start from c being with . So get
So which maps to, under , ( in the homology gamma's cancel as )
We now will move to an approach where we prove that our defined homology satisfies a series of established homology axioms that will allow us to apply the machinery of general homology to our specific "singular" homology defined via simplexes.
Axiom 0) Homology if a function
Definition The "category of chain complexes" is a category whose objects are chain complexes (of abelian groups) and morphisms which is a homomorphism between each abelian group in one chain and the corresponding group in the other chain such that the resulting diagram commutes. I.e,
Now, in our case, the chain complexes do in fact commute because is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity.
Homology of chain complexes is a functor in the natural way. That is, if for each p induces the functor
The proof is by "diagram chasing". Well, let , .
Let . Now . Furthermore, suppose . Then, so therefore some . This shows is well defined.
Thus, for get via the well defined functor .
1) Homotopy Axioms
If are homotopic then
Applications: If X and Y are homotopy equivalent then
let , such that and . Well and
Hence, and are invertible maps of each other. Q.E.D
Two morphisms between chain complexes are homotopic if you can find maps such that
Given H a homotopy connecting f Y we can construct a chain homotopy between Y)
If are chain homotopic then they induce equal maps on homology.
Proof of 2
Assume , that is,
(as and homology ignores exact forms)
Hence, at the level of homology they are the same.
Proof of 1
Consider a simplex in X. Now consider its image, a simplex, in Y under g and f respectively. Because of the homotopy we can construct a triangular based cylinder in Y with the image under f at the top and the image under y at the bottom.
Define = the above prism formed by and the homotopy H.
This, pictorially is correct but we need to be able to break up the prism, into a union of images of simplexes.
Suppose p=0, i.e. a point. Hence is a line, which is a simplex.
Suppose p=1 which yields a square. Adding a diagonal divides the square into two triangles, so is clearly a union of simplexes.
Suppose p=2. We get a prism which has a triangle for a base and a top. Raise each vertex on the bottom to the top in turn. This makes the prism a union of three simplexes.
In general for let and for vertexes
which is in
So have maps
Loosely, cuts each between the f_i and g_i and then deletes an entry. h\partial however does these in reverse order. Hence all that we are left with is