0708-1300/Class notes for Thursday, January 17: Difference between revisions
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'''Van-Kampen's Theorem''' |
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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. |
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Then, <math>\pi_1() = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)</math> |
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<math>\begin{matrix} |
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&\ \ \ \ U_1&&\\ |
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&\nearrow^{i_1}&\searrow^{j_1}&\\ |
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U_1\cap U_2&&&U_1\cup U_2 = X\\ |
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&\searrow_{i_2}&\nearrow^{j_2}&\\ |
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&\ \ \ \ U_2&&\\ |
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\end{matrix}</math> |
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where all the i's and j's are inclusions. |
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Lets consider the image of this under the functor <math>\pi_1</math> |
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<math>\begin{matrix} |
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&\ \ \ \ \pi_1(U_1)&&\\ |
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&\nearrow^{i_{1*}}&\searrow^{j_{1*}}&\\ |
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\pi_1(U_1\cap U_2)&&& \pi_1(X)\\ |
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&\searrow_{i_{2*}}&\nearrow^{j_{2*}}&\\ |
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&\ \ \ \ \pi(U_2)&&\\ |
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\end{matrix}</math> |
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Now consider the situation as groups: |
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<math>\begin{matrix} |
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&\ \ \ \ G_1&&\\ |
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&\nearrow_{\varphi_1}&\searrow&\\ |
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H&&&G_1*_H G_2\\ |
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&\searrow_{\varphi_2}&\nearrow&\\ |
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&\ \ \ \ G_2&&\\ |
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\end{matrix}</math> |
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Where <math>G_1 *_H G_2 = </math>{ words with letters alternating between being in <math>G_1</math> and <math>G_2</math>, ignoring e } / See Later |
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Considering just the set without the identification, we note this is a group with the operation being concatenation of words followed by reduction. |
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Ex: <math>a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4</math> where <math>a = a_2a_3</math> |
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'''Claim:''' |
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This is really a group. |
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So far, we have only defined the "free group of <math>G_1</math> and <math>G_2</math>". We now consider the identification (denoted above by 'See Later') which is |
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<math>\forall h\in H, \phi_1(h) = \phi_2(h</math>) |
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With this identification we have properly defined <math>G_1 *_H G_2</math> |
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Note: <math>G_1 *_H G_2</math> is equivalent to { words in <math>G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh)</math> |
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'''Example 0''' |
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<math>\pi_1(S^n) </math> for <math> n\geq 2</math> |
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We can think of <math>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>n\geq 2</math> this is connected (but fails for <math>S^1</math>) |
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So, <math>\pi_1(S^n) = \pi_1(U_1)*_{\pi(U_1\cap U_2)}\pi_1(U_2)</math> |
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But, since the hemispheres themselves are contractible, <math>\pi_1(U_1) = \pi_1(U_2) = \{e\}</math> |
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Hence, <math>\pi_1(S^n) = \{e\}</math> |
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'''Example 1''' |
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Let us consider <math>\pi_1</math> of a a figure eight. Let <math>U_1</math> denote everything above a line slightly beneath the intersection and <math>U_2</math> everything below a line slightly above the intersection point. |
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Now both <math>U_1</math> and <math>U_2</math> are homotopically equivalent to a loop and so <math>\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><\alpha></math> and <math><\beta></math> respectively. |
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The intersection is an X, contractible to a point and so <math>\pi_1(U_1\cap U_2) = \{e\}</math> |
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So <math>\pi_1</math>(figure 8)<math> = <\alpha>*_{\{\}}<\beta> = F(\alpha,\beta)</math> the free group generated by <math>\alpha</math> and <math>\beta</math> |
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This is non abelian |
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'''Example 2''' |
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<math>\pi_1(\mathbb{T}^2)</math> |
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We consider <math>\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>U_1</math> as everything inside the larger square and <math>U_2</math> as everything outside the smaller square. |
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Clearly <math>U_1</math> is contractible, and hence <math>\pi_1(U_1) = \{e\}</math> |
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Now, the intersection of <math>U_1</math> and <math>U_2</math> is equivalent to an annulus and so <math>\pi_1(U_1\cap U_2) = \mathbb{Z} = <\gamma></math> where <math>\gamma</math> is just a loop in the annulus. |
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Now considering <math>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. |
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Hence <math>\pi_1(U_2) = F(\alpha, \beta)</math> as in example 1 |
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Hence, <math>\pi_1(\mathbb{T}^2) = \{e\}*F(\alpha,\beta)/(i_{1*}(\gamma) = i_{2*}(\gamma))</math> |
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Now, <math>i_{1*}(\gamma) = e</math> |
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and |
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<math>i_{2*}(\gamma) = \alpha\beta\alpha^{-1}\beta^{-1}</math> |
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I.e., <math>\pi_1(\mathbb{T}^2) = F(\alpha,\beta)/ e = \alpha\beta\alpha^{-1}\beta^{-1}</math> |
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<math> = F(\alpha,\beta)/(\alpha\beta = \beta\alpha)</math> |
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This is just the Free Abelian group on two symbols and, |
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<math>= \{\alpha^n\beta^m\} = \mathbb{Z}^2</math> |
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Hence, <math>\pi_1(\mathbb{T}^2) = \mathbb{Z}^2</math> |
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'''Example 3''' |
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The two holed torus: <math>\Sigma_2</math> |
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Consider the schematic for this surface, consising of an octagon with edges labeled <math>a_1,b_1,a_1^{-1},b_1^{-1},a_2,b_2,a_2^{-1},b_2^{-1}</math> |
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As in the previous example, consider two concentric circles inside the octagon. Let everything inside the larger circle be <math>U_1</math> and everything outside the smaller circle be <math>U_2</math>. |
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Clearly <math>\pi_1(U_1) = \{e\}</math> as before. |
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<math>\pi_1(U_1\cap U_2) = <\gamma></math> as before. |
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Now, <math>U_2</math> this times when doing the identifications looks like a clover (4 loops intersecting at one point) |
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Completely analogously to before, we see that <math>\pi_1(U_2) = F(\alpha_1, \beta_1, \alpha_2, \beta_2)</math> |
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Again, <math>i_{1*}(\gamma) = e</math> |
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<math>i_{2*}(\gamma) = \alpha_1\beta_1\alpha_1^{-1}\alpha_2\beta_2\alpha_1^{-1}\beta_2^{-1}</math> |
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Therefore, |
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<math>\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> |
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The ''abelianization'' of this group is |
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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> |
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Revision as of 23:17, 21 January 2008
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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 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 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)}
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 \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 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}
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 \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:
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 \begin{matrix} &\ \ \ \ G_1&&\\ &\nearrow_{\varphi_1}&\searrow&\\ H&&&G_1*_H G_2\\ &\searrow_{\varphi_2}&\nearrow&\\ &\ \ \ \ G_2&&\\ \end{matrix}}
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 G_1 *_H G_2 = }
{ words with letters alternating between being in 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 G_1}
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: 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 a_1b_1a_2 + a_3b_2a_4 = a_1b_1ab_2a_4} 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 a = a_2a_3}
Claim:
This is really a group.
So far, we have only defined the "free group 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 G_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 G_2}
". We now consider the identification (denoted above by 'See Later') which 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 \forall h\in H, \phi_1(h) = \phi_2(h} )
With this identification we have properly defined 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 G_1 *_H G_2}
Note: 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 G_1 *_H G_2}
is equivalent to { words in 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 G_1\cap G_2\}/ (e_1 = \{\}, e_2 = \{\}, g,h\in G_i, g\cdot h = gh)}
Example 0
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 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,
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 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} 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 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 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) = \{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 as everything inside the larger square 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} 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,
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 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 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 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 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}
