0708-1300/Class notes for Thursday, January 17: Difference between revisions
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'''Van-Kampen's Theorem''' |
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{{In| |
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n = 1 | |
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in = <nowiki><< KnotTheory`</nowiki>}} |
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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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<tt>Loading KnotTheory` version of January 13, 2008, 20:30:12.1353.<br> |
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Read more at http://katlas.org/wiki/KnotTheory.</tt> |
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Then, <math>\pi_1(X) = \pi_1(U_1)*_{\pi_1(U_1\cap U_2)}\pi_1(U_2)</math> |
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{{Graphics| |
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n = 2 | |
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in = <nowiki>TubePlot[TorusKnot[8, 3]]</nowiki> | |
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img= 0708-1300-T83.png}} |
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{{In| |
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n = 3 | |
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in = <nowiki>TC[r1_, t1_,r2_,t2_ ] := { |
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(r1 +r2 Cos[2Pi t2])Cos[2Pi t1], |
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(r1 +r2 Cos[2Pi t2])Sin[2Pi t1], |
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r2 Sin[2Pi t2] |
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};</nowiki>}} |
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{{In| |
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n = 4 | |
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in = <nowiki>InflatedTorus[p_, q_, b_] := ParametricPlot3D[ |
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TC[ |
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2, p t - q s, |
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1 + b(p^2 + q^2)s(1 - (p^2 + q^2)s), q t + p s |
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], |
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{t, 0, 1}, {s, 0, 1/(p^2 + q^2)}, |
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PlotPoints -> {6(p^2 + q^2) + 1, 7}, |
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DisplayFunction -> Identity |
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];</nowiki>}} |
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<math>\begin{matrix} |
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{{Graphics| |
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&\ \ \ \ U_1&&\\ |
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&\nearrow^{i_1}&\searrow^{j_1}&\\ |
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in = <nowiki>GraphicsArray[{{InflatedTorus[3,8,1], InflatedTorus[3,8,-1]}}]</nowiki> | |
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U_1\cap U_2&&&U_1\cup U_2 = X\\ |
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img= 0708-1300-InflatedTori.png | |
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&\searrow_{i_2}&\nearrow^{j_2}&\\ |
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width = 720px}} |
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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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In case someone might want diagrams for the examples above: |
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[[Image:0708-1300_notes_17-01-08c.jpg|200px]] |
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Latest revision as of 18:16, 6 February 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(X) = \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:
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 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}
, 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
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
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) } 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 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 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 . 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 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
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 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, 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,
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}
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:
