Difference between revisions of "07081300/Class notes for Thursday, January 17"
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Let X be a point pointed topological space such that <math>X = U_1\cup U_2</math> where <math>U_1</math> and <math>U_2</math> are open and the base point b is in the (connected) intersection.  Let X be a point pointed topological space such that <math>X = U_1\cup U_2</math> where <math>U_1</math> and <math>U_2</math> are open and the base point b is in the (connected) intersection.  
−  Then, <math>\pi_1() = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)</math>  +  Then, <math>\pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)</math> 
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<math>\pi_1^{ab}(\Sigma_2) = \pi_1(\Sigma_2)/ gh=hg = F.A.G (\alpha_1,\alpha_2,\beta_1,\beta^2) = \mathbb{Z}^4 \neq \mathbb{Z}^2</math>  <math>\pi_1^{ab}(\Sigma_2) = \pi_1(\Sigma_2)/ gh=hg = F.A.G (\alpha_1,\alpha_2,\beta_1,\beta^2) = \mathbb{Z}^4 \neq \mathbb{Z}^2</math>  
+  
+  
+  In case someone might want diagrams for the examples above:  
+  
+  [[Image:07081300_notes_170108c.jpg200px]] 
Latest revision as of 18:16, 6 February 2008

VanKampen's Theorem
Let X be a point pointed topological space such that where and are open and the base point b is in the (connected) intersection.
Then,
where all the i's and j's are inclusions.
Lets consider the image of this under the functor
Now consider the situation as groups:
Where { words with letters alternating between being in and , ignoring e } / See Later
Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction.
Ex: where
Claim:
This is really a group.
So far, we have only defined the "free group of and ". We now consider the identification (denoted above by 'See Later') which is
)
With this identification we have properly defined
Note: is equivalent to { words in
Example 0
for
We can think of as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as this is connected (but fails for )
So,
But, since the hemispheres themselves are contractible,
Hence,
Example 1
Let us consider of a a figure eight. Let denote everything above a line slightly beneath the intersection and everything below a line slightly above the intersection point.
Now both and are homotopically equivalent to a loop and so . We can think of these being the groups generated by a loop going around once, I.e., isomorphic to and respectively.
The intersection is an X, contractible to a point and so
So (figure 8) the free group generated by and
This is non abelian
Example 2
We consider in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define as everything inside the larger square and as everything outside the smaller square.
Clearly is contractible, and hence
Now, the intersection of and is equivalent to an annulus and so where is just a loop in the annulus.
Now considering , we note that each of the four outer corner points in the big square are identified, and when we identify edges we are left with something equivalent to a figure 8.
Hence as in example 1
Hence,
Now,
and
I.e.,
This is just the Free Abelian group on two symbols and,
Hence,
Example 3
The two holed torus:
Consider the schematic for this surface, consising of an octagon with edges labeled
As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be and everything outside the smaller circle be .
Clearly as before.
as before.
Now, this times when doing the identifications looks like a clover (4 loops intersecting at one point)
Completely analogously to before, we see that
Again,
Therefore,
The abelianization of this group is
In case someone might want diagrams for the examples above: