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==A Homology Theory is a Monster==
[[Image:0708-1300-AxiomsForHomology.png|thumb|center|540px|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 "'''S'''tate" or "'''P'''rove" or "'''A'''pply".

==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: <math>H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}</math>

From this we get the pth homology for a topological space <math>H_p(X)</math>

We have previously shown that

''1) <math>H_p(\cup X) = \oplus H_p(X_i)</math> for disjoint unions of spaces <math>X_i</math>

2) <math>H_p(pt) = \mathbb{Z}\delta_{p0}</math>

3) <math>H_0(connected) = \mathbb{Z}</math>

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

<math>\phi:[\gamma]_{\pi_1^{ab}}\mapsto[\gamma]_{H_1}</math>

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

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

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

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

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

''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 simplices.

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, <math>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\}</math>

Now, in our case, the chain complexes do in fact commute because <math>\partial</math> is defined by pre-composition but f is defined by post-composition. Hence, associativity of composition yields commutativity.

'''Claim'''

Homology of chain complexes is a functor in the natural way. That is, if <math>f_p:C_p\rightarrow D_p</math> for each p induces the functor <math>f_*:H_p(C_*)\rightarrow H_p(D_*)</math>

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

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

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

===Second Hour===

'''1) Homotopy Axioms'''

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

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

''Proof:''

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''

'''Definition'''

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_*)] =[(\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 into a union of images of simplexes.

Suppose

2008-02-01T20:21:41Z

Trefor:

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==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.

'''Proof of Van Kampen'''

Let <math>G_i = \pi_1(U_i)</math>

<math>H = \pi_1(Y_1\cap U_2)</math>

<math>G=\pi_1(U_1\cup U_2)</math>

We aim to show that <math>G=G_1*_H G_2</math>

Hence, we want to define two maps:

<math>\Phi:G_1*_H G_2\rightarrow G </math> and

<math>\Psi:G\rightarrow G_1*_H G_2</math>

such that they are inverses of each other.

Now, recall the commuting diagram:

<math>\begin{matrix}
&\ \ \ \ U_1&&\\
&\nearrow^{i_1}&\searrow^{j_1}&\\
U_1\cap U_2&&&U_1\cup U_2 = X\\
&\searrow_{i_2}&\nearrow^{j_2}&\\
&\ \ \ \ U_2&&\\
\end{matrix}</math>

Further, let <math>b_i</math> alternate between 1 and 2 for successive i's.

Hence we define <math>\Phi</math> via for <math>\alpha_i\in G_{b_i}</math>,

[<math>\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]</math>

Clearly this is well defined. We need to check the relations in
<math>G_1*_H G_2</math> indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after <math>\Phi</math> and the definition necessitates the third relation holds.

Now for <math>\Psi</math>:

Elements in G correspond with paths <math>\gamma</math> in <math>U_1\cup U_2</math>

On consider such a <math>\gamma</math>. The Lesbegue Lemma let us break <math>\gamma</math> up so <math>\gamma = \gamma_1\ldots\gamma_N</math> such that each <math>\gamma_i</math> is in just <math>G_{b_i}</math>

Now, <math>\gamma_i</math> does not go from base point to base point, so we can't consider it as a loop itself. Let <math>x_i</math> denote the endpoint of <math>\gamma_i</math> (and hence <math>x_{i-1}</math> is the beginning point) and further, <math>x_0 = b</math> the base point.

Choose paths <math>\eta_i</math> connecting <math>x_i</math> to b such that if <math>x_i\in U_{b_i}</math> then <math>\eta_i\sub U_{b_i}</math>

Hence, <math>\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots</math>

where each section is entirely in <math>G_{b_i}</math>. This above mapping is <math>\Psi</math>

We need to show that <math>\Psi</math> is well defined. I.e.,

1) <math>\Psi</math> is independent of the subdivision

2) <math>\Psi</math> is independent of choice of the <math>\eta_i</math>'s

3) If <math>\gamma_1\sim\gamma_2</math> then <math>\Psi(\gamma_1) = \Psi(\gamma_2)</math>

For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points.

For 2), Suppose <math>\eta_i</math> goes from the basepoint to <math>x_i</math>. Add a subdivision point y right beside <math>x_i</math> and remove the one at <math>x_i</math>. Then add again the original point <math>x_i</math>. The trick is that the new <math>\eta_i'</math> is the one used when adding y. The map from the new <math>x_i</math> to the basepoint is this <math>\eta_i'</math> with the infinitesimal connection between y and <math>x_i</math> added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from <math>x_i</math> to b now being <math>\eta_i'</math> instead of the original <math>\eta_i</math>

For 3) we consider the homotopy between <math>\gamma_1</math> and <math>\gamma_2</math> thought of as a square with <math>\gamma_1</math> on the bottom and <math>\gamma_2</math> on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in <math>U_{b_i}</math>

We further modify the homotopy H to a new homotopy <math>\tilde{H}</math> such that each grid point gets mapped to b.

This can be thought of as "pinching" each gird point and pulling it to b or, alternatively, as tossing a "handkerchief on a bed of nails"

If we let <math>\gamma_1</math> be broken in to section <math>\alpha_1,\ldots, \alpha_n</math> along the bottom then <math>\tilde{H}</math> first lifts <math>\alpha_1</math> to the top and right sides of the subsquare such that the gridpoint goes to b. <math>\tilde{H}</math> then moves the next square up in an analogous manner until we are at the top with <math>\gamma_2</math>.

0708-1300/Class notes for Thursday, January 24

2008-02-01T20:21:31Z

Trefor: /* Typed Notes */

INCOMPLETE AND UNEDITED: Completion coming soon.

{{0708-1300/Navigation}}

==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.

'''Proof of Van Kampen'''

Let <math>G_i = \pi_1(U_i)</math>

<math>H = \pi_1(Y_1\cap U_2)</math>

<math>G=\pi_1(U_1\cup U_2)</math>

We aim to show that <math>G=G_1*_H G_2</math>

Hence, we want to define two maps:

<math>\Phi:G_1*_H G_2\rightarrow G </math> and

<math>\Psi:G\rightarrow G_1*_H G_2</math>

such that they are inverses of each other.

Now, recall the commuting diagram:

<math>\begin{matrix}
&\ \ \ \ U_1&&\\
&\nearrow^{i_1}&\searrow^{j_1}&\\
U_1\cap U_2&&&U_1\cup U_2 = X\\
&\searrow_{i_2}&\nearrow^{j_2}&\\
&\ \ \ \ U_2&&\\
\end{matrix}</math>

Further, let <math>b_i</math> alternate between 1 and 2 for successive i's.

Hence we define <math>\Phi</math> via for <math>\alpha_i\in G_{b_i}</math>,

[<math>\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]</math>

Clearly this is well defined. We need to check the relations in
<math>G_1*_H G_2</math> indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after <math>\Phi</math> and the definition necessitates the third relation holds.

Now for <math>\Psi</math>:

Elements in G correspond with paths <math>\gamma</math> in <math>U_1\cup U_2</math>

On consider such a <math>\gamma</math>. The Lesbegue Lemma let us break <math>\gamma</math> up so <math>\gamma = \gamma_1\ldots\gamma_N</math> such that each <math>\gamma_i</math> is in just <math>G_{b_i}</math>

Now, <math>\gamma_i</math> does not go from base point to base point, so we can't consider it as a loop itself. Let <math>x_i</math> denote the endpoint of <math>\gamma_i</math> (and hence <math>x_{i-1}</math> is the beginning point) and further, <math>x_0 = b</math> the base point.

Choose paths <math>\eta_i</math> connecting <math>x_i</math> to b such that if <math>x_i\in U_{b_i}</math> then <math>\eta_i\sub U_{b_i}</math>

Hence, <math>\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots</math>

where each section is entirely in <math>G_{b_i}</math>. This above mapping is <math>\Psi</math>

We need to show that <math>\Psi</math> is well defined. I.e.,

1) <math>\Psi</math> is independent of the subdivision

2) <math>\Psi</math> is independent of choice of the <math>\eta_i</math>'s

3) If <math>\gamma_1\sim\gamma_2</math> then <math>\Psi(\gamma_1) = \Psi(\gamma_2)</math>

For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points.

For 2), Suppose <math>\eta_i</math> goes from the basepoint to <math>x_i</math>. Add a subdivision point y right beside <math>x_i</math> and remove the one at <math>x_i</math>. Then add again the original point <math>x_i</math>. The trick is that the new <math>\eta_i'</math> is the one used when adding y. The map from the new <math>x_i</math> to the basepoint is this <math>\eta_i'</math> with the infinitesimal connection between y and <math>x_i</math> added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from <math>x_i</math> to b now being <math>\eta_i'</math> instead of the original <math>\eta_i</math>

For 3) we consider the homotopy between <math>\gamma_1</math> and <math>\gamma_2</math> thought of as a square with <math>\gamma_1</math> on the bottom and <math>\gamma_2</math> on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in <math>U_{b_i}</math>

We further modify the homotopy H to a new homotopy <math>\tilde{H}</math> such that each grid point gets mapped to b.

This can be thought of as "pinching" each gird point and pulling it to b or, alternatively, as tossing a "handkerchief on a bed of nails"

If we let <math>\gamma_1</math> be broken in to section <math>\alpha_1,\ldots, \alpha_n</math> along the bottom then <math>\tilde{H}</math> first lifts <math>\alpha_1</math> to the top and right sides of the subsquare such that the gridpoint goes to b. <math>\tilde{H}</math> then moves the next square up in an analogous manner until we are at the top with <math>\gamma_2</math>.