# 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 ${\displaystyle X=U_{1}\cup U_{2}}$ where ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ are open and the base point b is in the (connected) intersection.

Then, ${\displaystyle \pi _{1}(X)=\pi _{1}(U_{1})*_{\pi _{1}(U_{1}\cap U_{2})}\pi _{1}(U_{2})}$

${\displaystyle {\begin{matrix}&\ \ \ \ U_{1}&&\\&\nearrow ^{i_{1}}&\searrow ^{j_{1}}&\\U_{1}\cap U_{2}&&&U_{1}\cup U_{2}=X\\&\searrow _{i_{2}}&\nearrow ^{j_{2}}&\\&\ \ \ \ U_{2}&&\\\end{matrix}}}$

where all the i's and j's are inclusions.

Lets consider the image of this under the functor ${\displaystyle \pi _{1}}$

${\displaystyle {\begin{matrix}&\ \ \ \ \pi _{1}(U_{1})&&\\&\nearrow ^{i_{1*}}&\searrow ^{j_{1*}}&\\\pi _{1}(U_{1}\cap U_{2})&&&\pi _{1}(X)\\&\searrow _{i_{2*}}&\nearrow ^{j_{2*}}&\\&\ \ \ \ \pi (U_{2})&&\\\end{matrix}}}$

Now consider the situation as groups:

${\displaystyle {\begin{matrix}&\ \ \ \ G_{1}&&\\&\nearrow _{\varphi _{1}}&\searrow &\\H&&&G_{1}*_{H}G_{2}\\&\searrow _{\varphi _{2}}&\nearrow &\\&\ \ \ \ G_{2}&&\\\end{matrix}}}$

Where ${\displaystyle G_{1}*_{H}G_{2}=}${ words with letters alternating between being in ${\displaystyle G_{1}}$ and ${\displaystyle G_{2}}$, 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: ${\displaystyle a_{1}b_{1}a_{2}+a_{3}b_{2}a_{4}=a_{1}b_{1}ab_{2}a_{4}}$ where ${\displaystyle a=a_{2}a_{3}}$

Claim:

This is really a group.

So far, we have only defined the "free group of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2}}$". We now consider the identification (denoted above by 'See Later') which is

${\displaystyle \forall h\in H,\phi _{1}(h)=\phi _{2}(h}$)

With this identification we have properly defined ${\displaystyle G_{1}*_{H}G_{2}}$

Note: ${\displaystyle G_{1}*_{H}G_{2}}$ is equivalent to { words in ${\displaystyle G_{1}\cap G_{2}\}/(e_{1}=\{\},e_{2}=\{\},g,h\in G_{i},g\cdot h=gh)}$

Example 0

${\displaystyle \pi _{1}(S^{n})}$ for ${\displaystyle n\geq 2}$

We can think of ${\displaystyle S^{n}}$ as the union of two slightly overlapping open hemispheres which leaves the intersection as a band about the equator. As long as ${\displaystyle n\geq 2}$ this is connected (but fails for ${\displaystyle S^{1}}$)

So, ${\displaystyle \pi _{1}(S^{n})=\pi _{1}(U_{1})*_{\pi (U_{1}\cap U_{2})}\pi _{1}(U_{2})}$

But, since the hemispheres themselves are contractible, ${\displaystyle \pi _{1}(U_{1})=\pi _{1}(U_{2})=\{e\}}$

Hence, ${\displaystyle \pi _{1}(S^{n})=\{e\}}$

Example 1

Let us consider ${\displaystyle \pi _{1}}$ of a a figure eight. Let ${\displaystyle U_{1}}$ denote everything above a line slightly beneath the intersection and ${\displaystyle U_{2}}$ everything below a line slightly above the intersection point.

Now both ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ are homotopically equivalent to a loop and so ${\displaystyle \pi _{1}(U_{1})=\pi _{2}(U_{2})=\mathbb {Z} }$. We can think of these being the groups generated by a loop going around once, I.e., isomorphic to ${\displaystyle <\alpha >}$ and ${\displaystyle <\beta >}$ respectively.

The intersection is an X, contractible to a point and so ${\displaystyle \pi _{1}(U_{1}\cap U_{2})=\{e\}}$

So ${\displaystyle \pi _{1}}$(figure 8)${\displaystyle =<\alpha >*_{\{\}}<\beta >=F(\alpha ,\beta )}$ the free group generated by ${\displaystyle \alpha }$ and ${\displaystyle \beta }$

This is non abelian

Example 2

${\displaystyle \pi _{1}(\mathbb {T} ^{2})}$

We consider ${\displaystyle \mathbb {T} ^{2}}$ in the normal way as a square with the normal identifications on the sides. We then consider two concentric squares inside this and define ${\displaystyle U_{1}}$ as everything inside the larger square and ${\displaystyle U_{2}}$ as everything outside the smaller square.

Clearly ${\displaystyle U_{1}}$ is contractible, and hence ${\displaystyle \pi _{1}(U_{1})=\{e\}}$

Now, the intersection of ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ is equivalent to an annulus and so ${\displaystyle \pi _{1}(U_{1}\cap U_{2})=\mathbb {Z} =<\gamma >}$ where ${\displaystyle \gamma }$ is just a loop in the annulus.

Now considering ${\displaystyle U_{2}}$, 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 ${\displaystyle \pi _{1}(U_{2})=F(\alpha ,\beta )}$ as in example 1

Hence, ${\displaystyle \pi _{1}(\mathbb {T} ^{2})=\{e\}*F(\alpha ,\beta )/(i_{1*}(\gamma )=i_{2*}(\gamma ))}$

Now, ${\displaystyle i_{1*}(\gamma )=e}$

and

${\displaystyle i_{2*}(\gamma )=\alpha \beta \alpha ^{-1}\beta ^{-1}}$

I.e., ${\displaystyle \pi _{1}(\mathbb {T} ^{2})=F(\alpha ,\beta )/e=\alpha \beta \alpha ^{-1}\beta ^{-1}}$

${\displaystyle =F(\alpha ,\beta )/(\alpha \beta =\beta \alpha )}$

This is just the Free Abelian group on two symbols and,

${\displaystyle =\{\alpha ^{n}\beta ^{m}\}=\mathbb {Z} ^{2}}$

Hence, ${\displaystyle \pi _{1}(\mathbb {T} ^{2})=\mathbb {Z} ^{2}}$

Example 3

The two holed torus: ${\displaystyle \Sigma _{2}}$

Consider the schematic for this surface, consising of an octagon with edges labeled ${\displaystyle a_{1},b_{1},a_{1}^{-1},b_{1}^{-1},a_{2},b_{2},a_{2}^{-1},b_{2}^{-1}}$

As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be ${\displaystyle U_{1}}$ and everything outside the smaller circle be ${\displaystyle U_{2}}$.

Clearly ${\displaystyle \pi _{1}(U_{1})=\{e\}}$ as before.

${\displaystyle \pi _{1}(U_{1}\cap U_{2})=<\gamma >}$ as before.

Now, ${\displaystyle U_{2}}$ this times when doing the identifications looks like a clover (4 loops intersecting at one point)

Completely analogously to before, we see that ${\displaystyle \pi _{1}(U_{2})=F(\alpha _{1},\beta _{1},\alpha _{2},\beta _{2})}$

Again, ${\displaystyle i_{1*}(\gamma )=e}$

${\displaystyle i_{2*}(\gamma )=\alpha _{1}\beta _{1}\alpha _{1}^{-1}\alpha _{2}\beta _{2}\alpha _{1}^{-1}\beta _{2}^{-1}}$

Therefore,

${\displaystyle \pi _{\Sigma _{2}}=F(\alpha _{1},\beta _{1},\alpha _{2},\beta _{2})/(e=\alpha _{1}\beta _{1}\alpha _{1}^{-1}\alpha _{2}\beta _{2}\alpha _{1}^{-1}\beta _{2}^{-1})}$

The abelianization of this group is

${\displaystyle \pi _{1}^{ab}(\Sigma _{2})=\pi _{1}(\Sigma _{2})/gh=hg=F.A.G(\alpha _{1},\alpha _{2},\beta _{1},\beta ^{2})=\mathbb {Z} ^{4}\neq \mathbb {Z} ^{2}}$

In case someone might want diagrams for the examples above: