0708-1300/Class notes for Thursday, January 17
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Van-Kampen's Theorem
Let X be a point pointed topological space such that [math]\displaystyle{ X = U_1\cup U_2 }[/math] where [math]\displaystyle{ U_1 }[/math] and [math]\displaystyle{ U_2 }[/math] are open and the base point b is in the (connected) intersection.
Then, [math]\displaystyle{ \pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2) }[/math]
[math]\displaystyle{ \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]
where all the i's and j's are inclusions.
Lets consider the image of this under the functor [math]\displaystyle{ \pi_1 }[/math]
[math]\displaystyle{ \begin{matrix}
&\ \ \ \ \pi_1(U_1)&&\\
&\nearrow^{i_{1*}}&\searrow^{j_{1*}}&\\
\pi_1(U_1\cap U_2)&&& \pi_1(X)\\
&\searrow_{i_{2*}}&\nearrow^{j_{2*}}&\\
&\ \ \ \ \pi(U_2)&&\\
\end{matrix} }[/math]
Now consider the situation as groups:
[math]\displaystyle{ \begin{matrix}
&\ \ \ \ G_1&&\\
&\nearrow_{\varphi_1}&\searrow&\\
H&&&G_1*_H G_2\\
&\searrow_{\varphi_2}&\nearrow&\\
&\ \ \ \ G_2&&\\
\end{matrix} }[/math]
Where [math]\displaystyle{ G_1 *_H G_2 = }[/math]{ words with letters alternating between being in [math]\displaystyle{ G_1 }[/math] and [math]\displaystyle{ G_2 }[/math], 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: [math]\displaystyle{ a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4 }[/math] where [math]\displaystyle{ a = a_2a_3 }[/math]
Claim:
This is really a group.
So far, we have only defined the "free group of [math]\displaystyle{ G_1 }[/math] and [math]\displaystyle{ G_2 }[/math]". We now consider the identification (denoted above by 'See Later') which is
[math]\displaystyle{ \forall h\in H, \phi_1(h) = \phi_2(h }[/math])
With this identification we have properly defined [math]\displaystyle{ G_1 *_H G_2 }[/math]
Note: [math]\displaystyle{ G_1 *_H G_2 }[/math] is equivalent to { words in [math]\displaystyle{ G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh) }[/math]
Example 0
[math]\displaystyle{ \pi_1(S^n) }[/math] for [math]\displaystyle{ n\geq 2 }[/math]
We can think of [math]\displaystyle{ S^n }[/math] as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as [math]\displaystyle{ n\geq 2 }[/math] this is connected (but fails for [math]\displaystyle{ S^1 }[/math])
So, [math]\displaystyle{ \pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2) }[/math]
But, since the hemispheres themselves are contractible, [math]\displaystyle{ \pi_1(U_1) = \pi_1(U_2) = \{e\} }[/math]
Hence, [math]\displaystyle{ \pi_1(S^n) = \{e\} }[/math]
Example 1
Let us consider [math]\displaystyle{ \pi_1 }[/math] of a a figure eight. Let [math]\displaystyle{ U_1 }[/math] denote everything above a line slightly beneath the intersection and [math]\displaystyle{ U_2 }[/math] everything below a line slightly above the intersection point.
Now both [math]\displaystyle{ U_1 }[/math] and [math]\displaystyle{ U_2 }[/math] are homotopically equivalent to a loop and so [math]\displaystyle{ \pi_1(U_1) = \pi_2(U_2) = \mathbb{Z} }[/math]. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to [math]\displaystyle{ \lt \alpha\gt }[/math] and [math]\displaystyle{ \lt \beta\gt }[/math] respectively.
The intersection is an X, contractible to a point and so [math]\displaystyle{ \pi_1(U_1\cap U_2) = \{e\} }[/math]
So [math]\displaystyle{ \pi_1 }[/math](figure 8)[math]\displaystyle{ = \lt \alpha\gt *_{\{\}}\lt \beta\gt = F(\alpha,\beta) }[/math] the free group generated by [math]\displaystyle{ \alpha }[/math] and [math]\displaystyle{ \beta }[/math]
This is non abelian
Example 2
[math]\displaystyle{ \pi_1(\mathbb{T}^2) }[/math]
We consider [math]\displaystyle{ \mathbb{T}^2 }[/math] in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define [math]\displaystyle{ U_1 }[/math] as everything inside the larger square and [math]\displaystyle{ U_2 }[/math] as everything outside the smaller square.
Clearly [math]\displaystyle{ U_1 }[/math] is contractible, and hence [math]\displaystyle{ \pi_1(U_1) = \{e\} }[/math]
Now, the intersection of [math]\displaystyle{ U_1 }[/math] and [math]\displaystyle{ U_2 }[/math] is equivalent to an annulus and so [math]\displaystyle{ \pi_1(U_1\cap U_2) = \mathbb{Z} = \lt \gamma\gt }[/math] where [math]\displaystyle{ \gamma }[/math] is just a loop in the annulus.
Now considering [math]\displaystyle{ U_2 }[/math], 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 [math]\displaystyle{ \pi_1(U_2) = F(\alpha, \beta) }[/math] as in example 1
Hence, [math]\displaystyle{ \pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma)) }[/math]
Now, [math]\displaystyle{ i_{1*}(\gamma) = e }[/math]
and
[math]\displaystyle{ i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1} }[/math]
I.e., [math]\displaystyle{ \pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1} }[/math]
[math]\displaystyle{ = F(\alpha,\beta)/(\alpha\beta = \beta\alpha) }[/math]
This is just the Free Abelian group on two symbols and,
[math]\displaystyle{ = \{\alpha^n\beta^m\} = \mathbb{Z}^2 }[/math]
Hence, [math]\displaystyle{ \pi_1(\mathbb{T}^2) = \mathbb{Z}^2 }[/math]
Example 3
The two holed torus: [math]\displaystyle{ \Sigma_2 }[/math]
Consider the schematic for this surface, consising of an octagon with edges labeled [math]\displaystyle{ a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1} }[/math]
As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be [math]\displaystyle{ U_1 }[/math] and everything outside the smaller circle be [math]\displaystyle{ U_2 }[/math].
Clearly [math]\displaystyle{ \pi_1(U_1) = \{e\} }[/math] as before.
[math]\displaystyle{ \pi_1(U_1\cap U_2) = \lt \gamma\gt }[/math] as before.
Now, [math]\displaystyle{ U_2 }[/math] this times when doing the identifications looks like a clover (4 loops intersecting at one point)
Completely analogously to before, we see that [math]\displaystyle{ \pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2) }[/math]
Again, [math]\displaystyle{ i_{1*}(\gamma) = e }[/math]
[math]\displaystyle{ i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1} }[/math]
Therefore,
[math]\displaystyle{ \pi_{\Sigma_2} = F(\alpha_1, \beta_1, \alpha_2, \beta_2)/(e =\alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}) }[/math]
The abelianization of this group is
[math]\displaystyle{ \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:
