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Van-Kampen's Theorem

Let X be a point pointed topological space such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X = U_1\cup U_2} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} and $U_{2}$ are open and the base point b is in the (connected) intersection.

Then, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1() = \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\geq 2} this is connected (but fails for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S^1} )

So, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_1) = \pi_1(U_2) = \{e\}}

Hence, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(S^n) = \{e\}}

Example 1

Let us consider $\pi _{1}$ of a a figure eight. Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} denote everything above a line slightly beneath the intersection and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_2} everything below a line slightly above the intersection point.

Now both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_2} are homotopically equivalent to a loop and so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\alpha>} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <\beta>} respectively.

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

So Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1} (figure 8)Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = <\alpha>*_{\{\}}<\beta> = F(\alpha,\beta)} the free group generated by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta}

This is non abelian

Example 2

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(\mathbb{T}^2)}

We consider Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} as everything inside the larger square and $U_{2}$ as everything outside the smaller square.

Clearly Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} is contractible, and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_1) = \{e\}}

Now, the intersection of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_2} is equivalent to an annulus and so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_1\cap U_2) = \mathbb{Z} = <\gamma>} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma} is just a loop in the annulus.

Now considering Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_2) = F(\alpha, \beta)} as in example 1

Hence, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))}

Now, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i_{1*}(\gamma) = e}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}}

I.e., Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \{\alpha^n\beta^m\} = \mathbb{Z}^2}

Hence, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(\mathbb{T}^2) = \mathbb{Z}^2}

Example 3

The two holed torus: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_2} .

Clearly Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_1) = \{e\}} as before.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_1\cap U_2) = <\gamma>} as before.

Now, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)}

Again, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i_{1*}(\gamma) = e}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}}

Therefore,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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: