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Sep 10

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Errata to Bredon's Book


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VanKampen's Theorem
Let X be a point pointed topological space such that $X=U_{1}\cup U_{2}$ where $U_{1}$ and $U_{2}$ are open and the base point b is in the (connected) intersection.
Then, $\pi _{1}(X)=\pi _{1}(U_{1})*_{\pi _{1}(U_{1}\cap U_{2})}\pi _{1}(U_{2})$
${\begin{matrix}&\ \ \ \ U_{1}&&\\&\nearrow ^{i_{1}}&\searrow ^{j_{1}}&\\U_{1}\cap U_{2}&&&U_{1}\cup U_{2}=X\\&\searrow _{i_{2}}&\nearrow ^{j_{2}}&\\&\ \ \ \ U_{2}&&\\\end{matrix}}$
where all the i's and j's are inclusions.
Lets consider the image of this under the functor $\pi _{1}$
${\begin{matrix}&\ \ \ \ \pi _{1}(U_{1})&&\\&\nearrow ^{i_{1*}}&\searrow ^{j_{1*}}&\\\pi _{1}(U_{1}\cap U_{2})&&&\pi _{1}(X)\\&\searrow _{i_{2*}}&\nearrow ^{j_{2*}}&\\&\ \ \ \ \pi (U_{2})&&\\\end{matrix}}$
Now consider the situation as groups:
${\begin{matrix}&\ \ \ \ G_{1}&&\\&\nearrow _{\varphi _{1}}&\searrow &\\H&&&G_{1}*_{H}G_{2}\\&\searrow _{\varphi _{2}}&\nearrow &\\&\ \ \ \ G_{2}&&\\\end{matrix}}$
Where $G_{1}*_{H}G_{2}=${ words with letters alternating between being in $G_{1}$ and $G_{2}$, 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: $a_{1}b_{1}a_{2}+a_{3}b_{2}a_{4}=a_{1}b_{1}ab_{2}a_{4}$ where $a=a_{2}a_{3}$
Claim:
This is really a group.
So far, we have only defined the "free group of $G_{1}$ and $G_{2}$". We now consider the identification (denoted above by 'See Later') which is
$\forall h\in H,\phi _{1}(h)=\phi _{2}(h$)
With this identification we have properly defined $G_{1}*_{H}G_{2}$
Note: $G_{1}*_{H}G_{2}$ is equivalent to { words in $G_{1}\cap G_{2}\}/(e_{1}=\{\},e_{2}=\{\},g,h\in G_{i},g\cdot h=gh)$
Example 0
$\pi _{1}(S^{n})$ for $n\geq 2$
We can think of $S^{n}$ as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as $n\geq 2$ this is connected (but fails for $S^{1}$)
So, $\pi _{1}(S^{n})=\pi _{1}(U_{1})*_{\pi (U_{1}\cap U_{2})}\pi _{1}(U_{2})$
But, since the hemispheres themselves are contractible, $\pi _{1}(U_{1})=\pi _{1}(U_{2})=\{e\}$
Hence, $\pi _{1}(S^{n})=\{e\}$
Example 1
Let us consider $\pi _{1}$ of a a figure eight. Let $U_{1}$ denote everything above a line slightly beneath the intersection and $U_{2}$ everything below a line slightly above the intersection point.
Now both $U_{1}$ and $U_{2}$ are homotopically equivalent to a loop and so $\pi _{1}(U_{1})=\pi _{2}(U_{2})=\mathbb {Z}$. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to $<\alpha >$ and $<\beta >$ respectively.
The intersection is an X, contractible to a point and so $\pi _{1}(U_{1}\cap U_{2})=\{e\}$
So $\pi _{1}$(figure 8)$=<\alpha >*_{\{\}}<\beta >=F(\alpha ,\beta )$ the free group generated by $\alpha$ and $\beta$
This is non abelian
Example 2
$\pi _{1}(\mathbb {T} ^{2})$
We consider $\mathbb {T} ^{2}$ in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define $U_{1}$ as everything inside the larger square and $U_{2}$ as everything outside the smaller square.
Clearly $U_{1}$ is contractible, and hence $\pi _{1}(U_{1})=\{e\}$
Now, the intersection of $U_{1}$ and $U_{2}$ is equivalent to an annulus and so $\pi _{1}(U_{1}\cap U_{2})=\mathbb {Z} =<\gamma >$ where $\gamma$ is just a loop in the annulus.
Now considering $U_{2}$, 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 $\pi _{1}(U_{2})=F(\alpha ,\beta )$ as in example 1
Hence, $\pi _{1}(\mathbb {T} ^{2})=\{e\}*F(\alpha ,\beta )/(i_{1*}(\gamma )=i_{2*}(\gamma ))$
Now, $i_{1*}(\gamma )=e$
and
$i_{2*}(\gamma )=\alpha \beta \alpha ^{1}\beta ^{1}$
I.e., $\pi _{1}(\mathbb {T} ^{2})=F(\alpha ,\beta )/e=\alpha \beta \alpha ^{1}\beta ^{1}$
$=F(\alpha ,\beta )/(\alpha \beta =\beta \alpha )$
This is just the Free Abelian group on two symbols and,
$=\{\alpha ^{n}\beta ^{m}\}=\mathbb {Z} ^{2}$
Hence, $\pi _{1}(\mathbb {T} ^{2})=\mathbb {Z} ^{2}$
Example 3
The two holed torus: $\Sigma _{2}$
Consider the schematic for this surface, consising of an octagon with edges labeled $a_{1},b_{1},a_{1}^{1},b_{1}^{1},a_{2},b_{2},a_{2}^{1},b_{2}^{1}$
As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be $U_{1}$ and everything outside the smaller circle be $U_{2}$.
Clearly $\pi _{1}(U_{1})=\{e\}$ as before.
$\pi _{1}(U_{1}\cap U_{2})=<\gamma >$ as before.
Now, $U_{2}$ this times when doing the identifications looks like a clover (4 loops intersecting at one point)
Completely analogously to before, we see that $\pi _{1}(U_{2})=F(\alpha _{1},\beta _{1},\alpha _{2},\beta _{2})$
Again, $i_{1*}(\gamma )=e$
$i_{2*}(\gamma )=\alpha _{1}\beta _{1}\alpha _{1}^{1}\alpha _{2}\beta _{2}\alpha _{1}^{1}\beta _{2}^{1}$
Therefore,
$\pi _{\Sigma _{2}}=F(\alpha _{1},\beta _{1},\alpha _{2},\beta _{2})/(e=\alpha _{1}\beta _{1}\alpha _{1}^{1}\alpha _{2}\beta _{2}\alpha _{1}^{1}\beta _{2}^{1})$
The abelianization of this group is
$\pi _{1}^{ab}(\Sigma _{2})=\pi _{1}(\Sigma _{2})/gh=hg=F.A.G(\alpha _{1},\alpha _{2},\beta _{1},\beta ^{2})=\mathbb {Z} ^{4}\neq \mathbb {Z} ^{2}$
In case someone might want diagrams for the examples above: