Difference between revisions of "07081300/Class notes for Tuesday, February 26"
(→A Homology Theory is a Monster) 

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We have previously shown that  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>  +  ''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>  2) <math>H_p(pt) = \mathbb{Z}\delta_{p0}</math>  
Line 48:  Line 48:  
''Axiomized Homology''  ''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  +  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  Axiom 0) Homology if a function  
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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>.  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 <math>f,g:X\rightarrow Y</math> are homotopic then <math>f_* = g_*: H_p(X)\rightarrow H_p(Y)</math>  
+  
+  
+  Applications: If X and Y are homotopy equivalent then <math>H_*(X) \cong H_*(Y)</math>  
+  
+  ''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>  
+  
+  Hence, <math>f_*</math> and <math>g_*</math> are invertible maps of each other. ''Q.E.D''  
+  
+  
+  '''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_{p1}\partial_p</math>  
+  
+  
+  '''Claim 1'''  
+  
+  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)  
+  
+  
+  '''Claim 2'''  
+  
+  If <math>f,g:C_*\rightarrow C_*</math> are chain homotopic then they induce equal maps on homology.  
+  
+  
+  
+  ''Proof of 2''  
+  
+  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)  
+  
+  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 <math>h\sigma</math> = the above prism formed by <math>\sigma</math> and the homotopy H.  
+  
+  <math>(f_*g_*)\sigma = h\partial\sigma = \partial h\sigma</math>  
+  
+  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.  
+  
+  Suppose p=0, i.e. a point. Hence <math>\Delta_0\times I</math> 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 <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>  
+  
+  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>  
+  
+  So have maps <math>Y\leftarrow_H X\times I\leftarrow \Delta_p\times I \leftarrow\Delta_{p+1}</math>  
+  
+  
+  ''Claim: ''  
+  
+  <math>\partial h +h\partial = fg</math>  
+  
+  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> 
Revision as of 23:28, 7 March 2008

Contents 
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".
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:
From this we get the pth homology for a topological space
We have previously shown that
1) for disjoint unions of spaces
2)
3)
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 )
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 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 precomposition but f is defined by postcomposition. Hence, associativity of composition yields commutativity.
Claim
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 .
Second Hour
1) Homotopy Axioms
If are homotopic then
Applications: If X and Y are homotopy equivalent then
Proof:
let , such that and . Well and
Hence, and are invertible maps of each other. Q.E.D
Definition
Two morphisms between chain complexes are homotopic if you can find maps such that
Claim 1
Given H a homotopy connecting f Y we can construct a chain homotopy between Y)
Claim 2
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
Then,
which is in
So have maps
Claim:
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