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NOTE: This page is used as a placeholder for incomplete typed class notes.
 
  
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==Typed Notes==
  
{{0708-1300/Navigation}}
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<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>
  
==A Homology Theory is a Monster==
+
===First Hour===
[[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.
+
Recall we had defined <math>\tilde{H}(X):= ker\epsilon_*</math> where <math>\epsilon_*:X\rightarrow\{pt\}</math>, <math>\tilde{H}(X,A) = H(X,A)</math>
  
'''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".
+
For <math>X\neq\empty</math>, <math>\tilde{H}_p(X) = \tilde{H}_p(X)</math> for <math>p\neq 0</math> and equals <math>G\oplus\tilde{H}_0(X)</math> for <math>p=0</math>
  
==Typed Notes==
 
  
<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>
+
This homology definition satisfies the axioms with the following changes: Exactness only for <math>X\neq\empty</math> and the dimension axiom being <math>\tilde{H}_*(pt) = 0</math>. Furthermore, instead of additivity we have, under mild conditions of <math>b_0\in X</math> and <math>b_1\in Y</math> (ie non empty) define <math>X\vee Y:= X\cup Y / b_0\sim b_1</math> for a disjoint union. Then, <math>\tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)
 +
</math>
  
===First Hour===
+
We can actually get the above isomorphism in the following way. There are natural projection maps <math>p_x</math> and <math>p_y</math> from <math>X\vee Y</math> to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps <math>i_x</math> and <math>i_y</math> going the other way.  Then, <math>(p_{x*},p_{y*})</math> and <math>i_{x*}+i_{y*}</math> are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it.
  
 +
<math>\tilde{H}</math> is "kinda" natural:
  
 +
We have a chain complex where <math>C_p = <\sigma:\Delta_p\rightarrow X></math> where <math>\Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}</math>
  
Recall we had defined for a chain complex the associated homology groups: <math>H_p(C_*) :=\ ker\partial_p/im\partial_{p+1}</math>
+
We thus get that <math>\Delta_{-1} = \empty</math> since  <math>\sum x_1 = 0\neq 1</math> vacuously.
  
From this we get the pth homology for a topological space <math>H_p(X)</math>
+
Therefore, <math>C_{-1}(X) = <\sigma:\Delta_{-1}\rightarrow X> = \mathbb{Z}</math>
  
We have previously shown that
+
So, <math>\tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)</math>
  
''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>
+
Note: We have never actually specified that p is positive axiomatically. In fact, <math>\tilde{H}_p(S^n) = G</math> for p=n and 0 for <math>p\neq n</math> works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p.
  
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>
+
'''Degrees'''
  
<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.
+
<math>(G=\mathbb{Z})</math>
  
 +
<math>f:S^n\rightarrow S^n</math> then get <math>f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}
 +
</math>
  
We need to check the maps are in fact inverses of each other.
+
We thus define:
 +
<math>deg(f):=d = f_*(1)</math>
  
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.
+
<math>
 +
f_{1,2}:S^0\rightarrow S^0</math> has
  
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>
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<math>1) f_1 = I</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>
+
2) <math>f_2</math> = flip, ie <math>x_0\rightarrow -x_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>)
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<math>deg f_1 = deg I = 1</math> in all dimensions
  
 +
deg <math>f_2</math> = deg flip = -1
  
''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.
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'''Proposition'''
  
Axiom 0) Homology if a function
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Let <math>f:S^n\rightarrow S^n</math> be <math>x_0\mapsto-x_0</math> and <math>x_i\mapsto x_i</math> for i>0
  
'''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>
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then def f= -1
  
 +
''Proof:''
  
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.
+
We get two rows of the following sequence, with the induced maps from f going vertically between them:
  
 +
<math>\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)</math>
  
'''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 resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism <math>\partial</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.
+
'''Propositions:'''
 +
 
 +
1) if <math>f\sim g: S^n\rightarrow S^n</math> then deg f = deg g
 +
 
 +
2) <math>S^n\rightarrow^f s^n\rightarrow^g S^n</math> then <math>deg (g\circ f) = deg (f) deg (g)
 +
</math>
 +
 
 +
3) deg a where a is the antipodal map <math>x\mapsto -x</math> has <math>deg a = (-1)^{n+1}</math> on <math>S^n</math>
  
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===
 
===Second Hour===
  
'''1) Homotopy Axioms'''
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'''Corollary'''
  
If <math>f,g:X\rightarrow Y</math> are homotopic then <math>f_* = g_*: H_p(X)\rightarrow H_p(Y)</math>
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If n is even, a is not homotopic to I
  
  
Applications: If X and Y are homotopy equivalent then <math>H_*(X) \cong H_*(Y)</math>
+
'''Corollary''' Every <math>f:S^2\rightarrow S^2</math> has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)
 +
 
 +
 
 +
''Proof''
 +
 
 +
Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. ''Q.E.D''
  
''Proof:''
 
  
let <math>f:X\rightarrow Y</math>, <math>g:Y\rightarrow X</math> such that <math>f\circ g\sim id_y</math> and <math>g\circ f \sim id_x</math>.  Well <math>f_*\circ g_* = id_{H(Y)}</math> and <math>g_*\circ f_* = id_{H(X)}</math>
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'''Corollary'''
  
Hence, <math>f_*</math> and <math>g_*</math> are invertible maps of each other. ''Q.E.D''
+
Every vector field on <math>S^{2n}</math> has a zero, i.e., "on earth there must be a windless points" or "you can't comb the hair on a coconut"
  
 +
''Proof'' A non zero vector field induces a homotopy of I to a which is impossible.
  
'''Definition'''
 
  
Two morphisms <math>f,g:C_*\rightarrow D_*</math> between chain complexes are ''homotopic'' if you can find maps <math>h_p:C_p\rightarrow D_{p+1}</math> such that <math>f_p - g_p = \partial_{p+1} h_p = h_{p-1}\partial_p</math>
+
'''Theorem'''
  
 +
If <math>f:S^n\rightarrow S^n \ni y_0</math> is smooth (and every map may be approximated by one) and <math>y_0\in S^n</math> is a regular value (which occurs almost everywhere by Sard's Theorem) and <math>f^{-1}(y_0) = \{x_1,\cdots, x_n\}</math> then deg f = <math>\sum_{j=1}^k \pm 1 = \sum</math> sign (det(<math>df_{x_i}</math>))
  
'''Claim 1'''
+
.ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near <math>x_i</math> with coordinates near <math>y_0</math> using an orientation preserving rotation of <math>S^n</math>
  
Given H a homotopy connecting f<math>,g:X\rightarrow</math> Y we can construct a chain homotopy between <math>f_*,g_*:C_*(X)\rightarrow C_*(</math>Y)
 
  
 +
''Examples''
  
'''Claim 2'''
+
1) <math>S^1\rightarrow S^1</math> via <math>z\mapsto z^k</math>
  
If <math>f,g:C_*\rightarrow C_*</math> are chain homotopic then they induce equal maps on homology.  
+
This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign.  
  
 +
So, <math>deg f = \sum_k +1 = k</math>
  
  
''Proof of 2''
+
2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the <math>deg f = +1 -1 +1-1\cdots = 0</math>
  
Assume <math>[c]\in H_p(C_*)</math>, that is, <math>\partial c =0</math>
 
  
<math>[f_* c] - [g_* c] = [(f_*-g_*)] =[(\partial h + h\partial)c] = 0</math> (as <math>\partial c = 0</math> and homology ignores exact forms)
+
''Proof of Theorem''
  
Hence, at the level of homology they are the same.
 
  
 +
1) Let <math>T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}</math> be linear and norm preserving, <math>T\in M_{n+1\times n+1}</math> and <math>T^{T}T = I</math>.  Then, deg T = det T
  
''Proof of 1''
+
Proof: Every rigid rotation is a product of reflections.
  
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 <math>h\sigma</math> = the above prism formed by <math>\sigma</math> and the homotopy H.  
+
2) Let <math>A:\mathbb{R}^N\rightarrow\mathbb{R}^n</math> be any <math>n\times n</math> non singular matrix so that <math>A(\infty)=\infty</math> so this induces a map <math>\tilde{A}:S^n\rightarrow S^n</math>. Then, <math>deg\tilde{A} = sign(det A)</math>
  
<math>(f_*-g_*)\sigma = h\partial\sigma = \partial h\sigma</math>
+
Proof: Gaussian elimination results in making A a product of "elementary matrices" which come in three types: A matrix with 1's along the diagonal except one diagonal entry being <math>\lambda</math>. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a <math>\lambda</math> in some non diagonal location.
  
This, pictorially is correct but we need to be able to break up the prism, <math>\Delta_p\times I</math> into a union of images of simplexes.  
+
The latter of these is clearly homotopic to the identity by simply turning the <math>\lambda</math> off.  
  
Suppose p=0, i.e. a point. Hence <math>\Delta_0\times I</math> is a line, which is a simplex.  
+
The middle of these is just a reflection. The former of these, if <math>\lambda>0</math> it is clearly homotopic to the identity. But if <math>\lambda<0</math> then it is homotopic to a reflection.  
  
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.
+
3) <math>f:\mathbb{R}^n\rightarrow\mathbb{R}^n</math> such that <math>f^{-1}(0) = 0</math>, <math>df|_0 = A</math> non singular, <math>f(\infty) = \infty</math> so f defines <math>\tilde{f}:S^n\rightarrow S^n</math> then deg f = sign(det <math>df|_0)</math>
  
In general for <math>\Delta_p\times I</math> let <math>f_i = (l_i,0)</math> and <math>g_i = (l_i,1)</math> for vertexes <math>l_i</math>
+
Proof: Consider for <math>t\geq 1</math>, <math>f_t(x) := tf(x/1)</math>. Then, <math>f_1 = f</math>, <math>f_{\infty}=A</math>. This is a homotopy as it makes good sense for <math>t\in[0,\infty]</math>. So, <math>deg f = deg\tilde{A}</math>
  
Then, <math>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]</math>
 
  
which is in <math>C_{p+1}(Y)</math>
+
4) All that remains to prove the theorem is the shift from (0,0) to <math>(x_0,y_0)</math> which induces a rotation at <math>df|_{x_0}</math>
  
So have maps <math>Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}</math>
 
  
  
''Claim:  ''
+
==Homologies with non trivial negative p's?==
  
<math>\partial h +h\partial = f-g</math>
+
In the axiomatic definition of homology there is no specification that the homology groups <math>H_p(X)</math> must be necessarily trivial for negative p's. That said, in the singular homology that we are developing using CW complexes we only get non trivial homologies for positive p's. The question was raised in class of when you could possibly have non trivial homology at negative p's.
  
Loosely, <math>\partial h</math> cuts each <math>[f_0\cdots f_i g_i g_{i+1}\cdots g_p]</math> 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 <math>[f_0\cdots f_p] - [g_0\cdots g_p]</math>
+
There are infact homology theories that DO have non trivial homology for negative p. In particular, Dr. Putnam from the University of Victoria has defined a homology theory on Smale space and this theory does in fact have non trivial homology groups at negative p's. While Smale spaces have a complicated axiomized definition, very loosely they are a topological space equipped with a metric and a homeomorphism  from the space to itself such that locally you can write the space as the direct sum of a section of the space that increases under the homeomorphism(in terms of the metric) and one that decreases. I.e., it is like you have a local coordinate system that tells you where the space is expanding and where it is contracting under the homeomorphism. The classic example of this are Shifts of Finite Type from Dynamical systems. I computed the homology of a particular Smale space consisting of a torus as the space and a particular homeomorphism of it, and found it to have a non trivial homology at <math>H_{-1}
 +
</math>

Latest revision as of 13:44, 18 March 2008

Announcements go here

Contents

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 \tilde{H}(X):= ker\epsilon_* where \epsilon_*:X\rightarrow\{pt\}, \tilde{H}(X,A) = H(X,A)

For X\neq\empty, \tilde{H}_p(X) = \tilde{H}_p(X) for p\neq 0 and equals G\oplus\tilde{H}_0(X) for p=0


This homology definition satisfies the axioms with the following changes: Exactness only for X\neq\empty and the dimension axiom being \tilde{H}_*(pt) = 0. Furthermore, instead of additivity we have, under mild conditions of b_0\in X and b_1\in Y (ie non empty) define X\vee Y:= X\cup Y / b_0\sim b_1 for a disjoint union. Then, \tilde{H}(X\vee Y)\cong \tilde{H}(X)\oplus \tilde{H}(Y)

We can actually get the above isomorphism in the following way. There are natural projection maps p_x and p_y from X\vee Y to X and Y respectively that simply contract Y and X respectively to the glued base point. There are also natural inclusion maps i_x and i_y going the other way. Then, (p_{x*},p_{y*}) and i_{x*}+i_{y*} are the two maps in the isomorphism. Proving they are in fact an isomorphism is a homework problem that uses excision to prove it.

\tilde{H} is "kinda" natural:

We have a chain complex where C_p = <\sigma:\Delta_p\rightarrow X> where \Delta_p=\{x:\mathbb{R}^{n+1}_{\geq 0}\ :\ \sum x_i = 1\}

We thus get that \Delta_{-1} = \empty since \sum x_1 = 0\neq 1 vacuously.

Therefore, C_{-1}(X) = <\sigma:\Delta_{-1}\rightarrow X> = \mathbb{Z}

So, \tilde{C}_*(X,A) = \tilde{C}_*(X)/\tilde{C}_*(A) = C(X,A)


Note: We have never actually specified that p is positive axiomatically. In fact, \tilde{H}_p(S^n) = G for p=n and 0 for p\neq n works fine for all p's. So, since the spaces we are going to be interested in are those that can be constructed from spheres we really will only encounter non trivial homologies for positive p.


Degrees

(G=\mathbb{Z})

f:S^n\rightarrow S^n then get f_*:\mathbb{Z}\cong \tilde{H}_n(S^n)\rightarrow \tilde{H}_n(S^n)\cong\mathbb{Z}

We thus define: deg(f):=d = f_*(1)


f_{1,2}:S^0\rightarrow S^0 has

1) f_1 = I

2) f_2 = flip, ie x_0\rightarrow -x_0

deg f_1 = deg I = 1 in all dimensions

deg f_2 = deg flip = -1


Proposition

Let f:S^n\rightarrow S^n be x_0\mapsto-x_0 and x_i\mapsto x_i for i>0

then def f= -1

Proof:

We get two rows of the following sequence, with the induced maps from f going vertically between them:

\tilde{H}_{n-1}(S^{n-1})\leftarrow^{\partial}\tilde{H}_n(D^n,S^{n-1})\rightarrow^{i_*} H_n(S^n, D^n_+)\leftarrow^{j_*}H_n(S^n)


The resulting diagram from the two rows of the above sequence and the maps induced by f between them in fact commute at all places, where the left square commutes as a result of the properties of the connecting homomorphism \partial


Propositions:

1) if f\sim g: S^n\rightarrow S^n then deg f = deg g

2) S^n\rightarrow^f s^n\rightarrow^g S^n then deg (g\circ f) = deg (f) deg (g)

3) deg a where a is the antipodal map x\mapsto -x has deg a = (-1)^{n+1} on S^n


Second Hour

Corollary

If n is even, a is not homotopic to I


Corollary Every f:S^2\rightarrow S^2 has a fixed point, or an antipodal point. Ie. f(x) = x or f(x) = -x for some value or x. (Note this is believed true for 2n not just 2, but the follow proof appears needs some modification to work in dimensions 2n)


Proof

Suppose f has no fixed points. Thus x and f(x) are distinct and define a great circle. Thus there is a shortest path from f(x) to -x. This uniquely defines a homotopy between f and a. Suppose f also had no antipodal points. Then the same great circle defines a unique homotopy between f and I. But I is not homotopic to a, a contradiction. Q.E.D


Corollary

Every vector field on S^{2n} has a zero, i.e., "on earth there must be a windless points" or "you can't comb the hair on a coconut"

Proof A non zero vector field induces a homotopy of I to a which is impossible.


Theorem

If f:S^n\rightarrow S^n \ni y_0 is smooth (and every map may be approximated by one) and y_0\in S^n is a regular value (which occurs almost everywhere by Sard's Theorem) and f^{-1}(y_0) = \{x_1,\cdots, x_n\} then deg f = \sum_{j=1}^k \pm 1 = \sum sign (det(df_{x_i}))

.ie. we get +1 if it preserves orientation and -1 if it reverses it. The latter term is done using an identification of the coordinates near x_i with coordinates near y_0 using an orientation preserving rotation of S^n


Examples

1) S^1\rightarrow S^1 via z\mapsto z^k

This map wraps the circle around itself k times yielding k preimages for each point in the image, all with the same sign.

So, deg f = \sum_k +1 = k


2) Consider the map of a sphere where you place a plastic bag over a sphere, collect the bag at a pole, twist it once, rewrap the sphere, twist and rewrap again k times. Then the deg f = +1 -1 +1-1\cdots = 0


Proof of Theorem


1) Let T:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1} be linear and norm preserving, T\in M_{n+1\times n+1} and T^{T}T = I. Then, deg T = det T

Proof: Every rigid rotation is a product of reflections.


2) Let A:\mathbb{R}^N\rightarrow\mathbb{R}^n be any n\times n non singular matrix so that A(\infty)=\infty so this induces a map \tilde{A}:S^n\rightarrow S^n. Then, deg\tilde{A} = sign(det A)

Proof: Gaussian elimination results in making A a product of "elementary matrices" which come in three types: A matrix with 1's along the diagonal except one diagonal entry being \lambda. A matrix which is the identity only with two rows interchanged. A matrix which is the identity with a \lambda in some non diagonal location.

The latter of these is clearly homotopic to the identity by simply turning the \lambda off.

The middle of these is just a reflection. The former of these, if \lambda>0 it is clearly homotopic to the identity. But if \lambda<0 then it is homotopic to a reflection.


3) f:\mathbb{R}^n\rightarrow\mathbb{R}^n such that f^{-1}(0) = 0, df|_0 = A non singular, f(\infty) = \infty so f defines \tilde{f}:S^n\rightarrow S^n then deg f = sign(det df|_0)

Proof: Consider for t\geq 1, f_t(x) := tf(x/1). Then, f_1 = f, f_{\infty}=A. This is a homotopy as it makes good sense for t\in[0,\infty]. So, deg f = deg\tilde{A}


4) All that remains to prove the theorem is the shift from (0,0) to (x_0,y_0) which induces a rotation at df|_{x_0}


Homologies with non trivial negative p's?

In the axiomatic definition of homology there is no specification that the homology groups H_p(X) must be necessarily trivial for negative p's. That said, in the singular homology that we are developing using CW complexes we only get non trivial homologies for positive p's. The question was raised in class of when you could possibly have non trivial homology at negative p's.

There are infact homology theories that DO have non trivial homology for negative p. In particular, Dr. Putnam from the University of Victoria has defined a homology theory on Smale space and this theory does in fact have non trivial homology groups at negative p's. While Smale spaces have a complicated axiomized definition, very loosely they are a topological space equipped with a metric and a homeomorphism from the space to itself such that locally you can write the space as the direct sum of a section of the space that increases under the homeomorphism(in terms of the metric) and one that decreases. I.e., it is like you have a local coordinate system that tells you where the space is expanding and where it is contracting under the homeomorphism. The classic example of this are Shifts of Finite Type from Dynamical systems. I computed the homology of a particular Smale space consisting of a torus as the space and a particular homeomorphism of it, and found it to have a non trivial homology at H_{-1}