Trefor: /* Typed Notes */

2008-02-10T02:15:41Z

Typed Notes

← Older revision		Revision as of 22:15, 9 February 2008
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	Let <math>G_i = \pi_1(U_i)</math>		Let <math>G_i = \pi_1(U_i)</math>

	<math>H = \pi_1(~~Y_1~~\cap U_2)</math>		<math>H = \pi_1(U_1\cap U_2)</math>

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

Trefor at 20:21, 1 February 2008

2008-02-01T20:21:41Z

← Older revision		Revision as of 16:21, 1 February 2008
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	INCOMPLETE AND UNEDITED: Completion coming soon.


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Trefor: /* Typed Notes */

2008-02-01T20:21:31Z

Typed Notes

@@ Line 11: / Line 11: @@
 '''Proof of Van Kampen'''
-Let G_i = \pi_1(U_i)
+Let <math>G_i = \pi_1(U_i)</math>
 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.
@@ Line 38: / Line 41: @@
-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
-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.

Trefor at 18:03, 28 January 2008

2008-01-28T18:03:25Z

New page

INCOMPLETE AND UNEDITED: Completion coming soon.

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==Typed Notes==

<span style="color: red;">The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.</span>

'''Proof of Van Kampen'''

Let G_i = \pi_1(U_i)
H = \pi_1(Y_1\cap U_2)
G=\pi_1(U_1\cup U_2)

We aim to show that G=G_1*_H G_2

Hence, we want to define two maps:

\Phi:G_1*_H G_2\rightarrow G and

\Psi:G\rightarrow G_1*_H G_2

such that they are inverses of each other.

Now, recall the commuting diagram:

<math>\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}</math>

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

Hence we define \Phi via for \alpha_i\in G_{b_i},

[\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]

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.

0708-1300/Class notes for Thursday, January 24 - Revision history

Trefor: /* Typed Notes */

Trefor at 20:21, 1 February 2008

Trefor: /* Typed Notes */

Trefor at 18:03, 28 January 2008