0708-1300/Class notes for Tuesday, February 26

From Drorbn
Revision as of 13:34, 26 April 2008 by Trefor (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
Announcements go here


A Homology Theory is a Monster

Page 183 of Bredon's book

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".

Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

First Hour

Recall we had defined for a chain complex the associated homology groups: H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}

From this we get the pth homology for a topological space H_p(X)

We have previously shown that

1) H_p(\cup X) = \oplus H_p(X_i) for disjoint unions of spaces X_i

2) H_p(pt) = \mathbb{Z}\delta_{p0}

3) H_0(connected) = \mathbb{Z}

4) H_1(connected) \cong \pi_1^{ab}(X) via the map


\psi:\sigma\in C_1\mapsto[\gamma_{\sigma(0)}\sigma\bar{\gamma_{\sigma(1)}}] where \sigma_x is a path connecting x_0 to x.

We need to check the maps are in fact inverses of each other.

Lets consider \psi\circ\phi. We start with a closed path starting at x_0 (thought of as in the fundamental group). \phi means we now think of it as a simplex in X with a point at x_0. \Psi now takes this to the path that parks at x_0 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 \phi\circ\psi. Start with just a path \sigma. Then \psi makes a loop adding two paths from the x_0 to the start and finish of \sigma forming a triangular like closed loop. We think of this loop as \sigma'\in C_1

Now, we start from c being c = \sum a_i\sigma_i with \partial c = 0. So get \sum a_i(\partial \sigma_1) = \sum a_i(\sigma_i(1)-\sigma_i(0))

So \Psi(c) = [\gamma_{\sigma_i(0)}\sigma_i\bar{\gamma_{\sigma_i(1)}}]_{\pi_1} which maps to, under \phi, \sum a_i(\gamma_{\sigma_i(0)} + \sigma_i - \gamma_{\sigma_i(1)}) = \sum a_i\sigma_i = c ( in the homology gamma's cancel as \partial c = 0)

Axiomized Homology

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 functor

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, Mor((C_p)_{p=0}^{\infty}, (D_p)^{\infty}_{p=0}) = \{(f_p:C_p\rightarrow D_p)_{p=0}^{\infty}\ |\ f_{p-1}\partial_p^C=\partial_p^D f_p\}

Now, in our case, the chain complexes do in fact commute because \partial 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 f_p:C_p\rightarrow D_p for each p induces the functor f_*:H_p(C_*)\rightarrow H_p(D_*)

The proof is by "diagram chasing". Well, let c\in C_p, \partial c =0.

Let f*[c] = [fc]. Now \partial fc  = f\partial c = 0. Furthermore, suppose c = \partial b. Then, f_*c = fc = \partial fb so therefore fc = \partial\beta some \beta. This shows f_* is well defined.

Thus, for c\in H_p(C_*) get fc\in H_p(D_*) via the well defined functor f_*.

Second Hour

1) Homotopy Axioms

If f,g:X\rightarrow Y are homotopic then f_* = g_*: H_p(X)\rightarrow H_p(Y)

Applications: If X and Y are homotopy equivalent then H_*(X) \cong H_*(Y)


let f:X\rightarrow Y, g:Y\rightarrow X such that f\circ g\sim id_y and g\circ f \sim id_x. Well f_*\circ g_* = id_{H(Y)} and g_*\circ f_* = id_{H(X)}

Hence, f_* and g_* are invertible maps of each other. Q.E.D


Two morphisms f,g:C_*\rightarrow D_* between chain complexes are homotopic if you can find maps h_p:C_p\rightarrow D_{p+1} such that f_p - g_p = \partial_{p+1} h_p + h_{p-1}\partial_p

Claim 1

Given H a homotopy connecting f,g:X\rightarrow Y we can construct a chain homotopy between f_*,g_*:C_*(X)\rightarrow C_*(Y)

Claim 2

If f,g:C_*\rightarrow C_* are chain homotopic then they induce equal maps on homology.

Proof of 2

Assume [c]\in H_p(C_*), that is, \partial c =0

[f_* c] - [g_* c] = [(f_*-g_*)c] =[(\partial h + h\partial)c] = 0 (as \partial c = 0 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 h\sigma = the above prism formed by \sigma and the homotopy H.

(f_*-g_*)\sigma = h\partial\sigma + \partial h\sigma

This, pictorially is correct but we need to be able to break up the prism, \Delta_p\times I into a union of images of simplexes.

Suppose p=0, i.e. a point. Hence \Delta_0\times I 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 \Delta_p\times I let f_i = (l_i,0) and g_i = (l_i,1) for vertexes l_i

Then, h\sigma = \sum_{i=0}^p (-1)^i H\circ(\sigma\times I)\circ[f_0\cdots f_i g_i g_{i+1}\cdots g_p]

which is in C_{p+1}(Y)

So have maps Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}


\partial h +h\partial = f-g

Loosely, \partial h cuts each [f_0\cdots f_i g_i g_{i+1}\cdots g_p] 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 [f_0\cdots f_p] - [g_0\cdots g_p]