0708-1300/Class notes for Thursday, January 24: Difference between revisions

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'''Proof of Van Kampen'''
'''Proof of Van Kampen'''


Let G_i = \pi_1(U_i)
Let <math>G_i = \pi_1(U_i)</math>
H = \pi_1(Y_1\cap U_2)
G=\pi_1(U_1\cup U_2)


<math>H = \pi_1(Y_1\cap U_2)</math>
We aim to show that G=G_1*_H G_2

<math>G=\pi_1(U_1\cup U_2)</math>

We aim to show that <math>G=G_1*_H G_2</math>




Hence, we want to define two maps:
Hence, we want to define two maps:


\Phi:G_1*_H G_2\rightarrow G and
<math>\Phi:G_1*_H G_2\rightarrow G </math> and



\Psi:G\rightarrow G_1*_H G_2
<math>\Psi:G\rightarrow G_1*_H G_2</math>


such that they are inverses of each other.
such that they are inverses of each other.
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Further, let b_i alternate between 1 and 2 for successive i's.
Further, let <math>b_i</math> alternate between 1 and 2 for successive i's.


Hence we define \Phi via for \alpha_i\in G_{b_i},
Hence we define <math>\Phi</math> via for <math>\alpha_i\in G_{b_i}</math>,


[\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]
[<math>\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]</math>


Clearly this is well defined. We need to check the relations in
Clearly this is well defined. We need to check the relations in
G_1*_H G_2 indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds.
<math>G_1*_H G_2</math> indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after <math>\Phi</math> and the definition necessitates the third relation holds.


Now for <math>\Psi</math>:

Elements in G correspond with paths <math>\gamma</math> in <math>U_1\cup U_2</math>


On consider such a <math>\gamma</math>. The Lesbegue Lemma let us break <math>\gamma</math> up so <math>\gamma = \gamma_1\ldots\gamma_N</math> such that each <math>\gamma_i</math> is in just <math>G_{b_i}</math>

Now, <math>\gamma_i</math> does not go from base point to base point, so we can't consider it as a loop itself. Let <math>x_i</math> denote the endpoint of <math>\gamma_i</math> (and hence <math>x_{i-1}</math> is the beginning point) and further, <math>x_0 = b</math> the base point.

Choose paths <math>\eta_i</math> connecting <math>x_i</math> to b such that if <math>x_i\in U_{b_i}</math> then <math>\eta_i\sub U_{b_i}</math>

Hence, <math>\gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots</math>

where each section is entirely in <math>G_{b_i}</math>. This above mapping is <math>\Psi</math>

We need to show that <math>\Psi</math> is well defined. I.e.,

1) <math>\Psi</math> is independent of the subdivision

2) <math>\Psi</math> is independent of choice of the <math>\eta_i</math>'s

3) If <math>\gamma_1\sim\gamma_2</math> then <math>\Psi(\gamma_1) = \Psi(\gamma_2)</math>


For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points.


For 2), Suppose <math>\eta_i</math> goes from the basepoint to <math>x_i</math>. Add a subdivision point y right beside <math>x_i</math> and remove the one at <math>x_i</math>. Then add again the original point <math>x_i</math>. The trick is that the new <math>\eta_i'</math> is the one used when adding y. The map from the new <math>x_i</math> to the basepoint is this <math>\eta_i'</math> with the infinitesimal connection between y and <math>x_i</math> added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from <math>x_i</math> to b now being <math>\eta_i'</math> instead of the original <math>\eta_i</math>


For 3) we consider the homotopy between <math>\gamma_1</math> and <math>\gamma_2</math> thought of as a square with <math>\gamma_1</math> on the bottom and <math>\gamma_2</math> on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in <math>U_{b_i}</math>

We further modify the homotopy H to a new homotopy <math>\tilde{H}</math> such that each grid point gets mapped to b.

This can be thought of as "pinching" each gird point and pulling it to b or, alternatively, as tossing a "handkerchief on a bed of nails"

If we let <math>\gamma_1</math> be broken in to section <math>\alpha_1,\ldots, \alpha_n</math> along the bottom then <math>\tilde{H}</math> first lifts <math>\alpha_1</math> to the top and right sides of the subsquare such that the gridpoint goes to b. <math>\tilde{H}</math> then moves the next square up in an analogous manner until we are at the top with <math>\gamma_2</math>.

Revision as of 15:21, 1 February 2008

INCOMPLETE AND UNEDITED: Completion coming soon.


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Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.


Proof of Van Kampen

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 G_i = \pi_1(U_i)}

We aim to show that


Hence, we want to define two maps:

and


such that they are inverses of each other.


Now, recall the commuting diagram:


Further, let alternate between 1 and 2 for successive i's.

Hence we define via for ,

[

Clearly this is well defined. We need to check the relations in indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after and the definition necessitates the third relation holds.


Now for :

Elements in G correspond with paths 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 U_1\cup U_2}


On consider such a 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} . The Lesbegue Lemma let us break 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} up 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 \gamma = \gamma_1\ldots\gamma_N} such that each 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_i} is in just 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_{b_i}}

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 \gamma_i} does not go from base point to base point, so we can't consider it as a loop itself. 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 x_i} denote the endpoint of (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 x_{i-1}} is the beginning point) and further, 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_0 = b} the base point.

Choose paths 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 \eta_i} connecting 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_i} to b such that if 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_i\in U_{b_i}} 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 \eta_i\sub U_{b_i}}

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 \gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots}

where each section is entirely 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_{b_i}} . This above mapping is

We need to show 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 \Psi} is well defined. I.e.,

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 \Psi} is independent of the subdivision

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 \Psi} is independent of choice of the 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 \eta_i} 's

3) If 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_1\sim\gamma_2} 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 \Psi(\gamma_1) = \Psi(\gamma_2)}


For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points.


For 2), Suppose 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 \eta_i} goes from the basepoint 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 x_i} . Add a subdivision point y right beside and remove the one at 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_i} . Then add again the original point 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_i} . The trick is that the new 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 \eta_i'} is the one used when adding y. The map from the new 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_i} to the basepoint is this 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 \eta_i'} with the infinitesimal connection between y 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 x_i} added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from 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_i} to b now being 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 \eta_i'} instead of the original 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 \eta_i}


For 3) we consider the homotopy between 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_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 \gamma_2} thought of as a square with 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_1} on the bottom 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 \gamma_2} on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely 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 U_{b_i}}

We further modify the homotopy H to a new homotopy such that each grid point gets mapped to b.

This can be thought of as "pinching" each gird point and pulling it to b or, alternatively, as tossing a "handkerchief on a bed of nails"

If we 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 \gamma_1} be broken in to section 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_1,\ldots, \alpha_n} along the bottom then first lifts to the top and right sides of the subsquare such that the gridpoint goes to b. 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 \tilde{H}} then moves the next square up in an analogous manner until we are at the top with 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_2} .