0708-1300/Class notes for Tuesday, January 15 - Revision history

Trefor: /* Second Hour */

2008-02-08T23:29:02Z

Second Hour

← Older revision		Revision as of 19:29, 8 February 2008
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	'''Theorem'''		'''Theorem'''

	<math>\pi_1(X+Y)\cong\pi_1(X)+\pi_1(Y)</math>		<math>\pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y)</math>

Trefor: /* Second Hour */

2008-02-08T23:27:01Z

Second Hour

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	Consider <math>X\rightarrow^f Y</math> and <math>Y\rightarrow^g X</math> forming a commuting diagram and lets consider its image under <math>\pi_1</math>		Consider <math>X\rightarrow^f Y</math> and <math>Y\rightarrow^g X</math> forming a commuting diagram and lets consider its image under <math>\pi_1</math>

	Then we get <math>\pi_1(X)\rightarrow^{f_*} \pi_1(Y)</math> and <math>\pi_1 (Y)\rightarrow^{~~g_8~~} \pi_1(X</math>) as a commuting diagram.		Then we get <math>\pi_1(X)\rightarrow^{f_} \pi_1(Y)</math> and <math>\pi_1 (Y)\rightarrow^{g_} \pi_1(X</math>) as a commuting diagram.

	We know that <math>f\circ g</math> is homotopic to the identity, and thus we also get <math>f_\circ g_</math> homotopic to the identity		We know that <math>f\circ g</math> is homotopic to the identity, and thus we also get <math>f_\circ g_</math> homotopic to the identity

Trefor: /* Second Hour */

2008-02-08T22:06:11Z

Second Hour

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	''Proof of Corollary:''		''Proof of Corollary:''

	Consider <math>f:S^2\rightarrow\mathbb{R}</math> by <math>p\mapsto</math> dist to <math>A_1</math> + dist to <math>A_2</math>		Consider <math>f:S^2\rightarrow\mathbb{R}^2</math> by <math>p\mapsto</math> (dist to <math>A_1</math>, dist to <math>A_2</math>)

	If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same <math>A_i</math>.		If f(p) = f(-p) then, for both possible cases of f(p) being zero or positive, we get that p and -p are in the same <math>A_i</math>.
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	Assume <math>f:S^2\rightarrow\mathbb{R}</math> has no p with f(p)=f(-p)		Assume <math>f:S^2\rightarrow\mathbb{R}^2</math> has no p with f(p)=f(-p)

	Define <math>g:S^2\rightarrow S^1</math> by <math>g(p) = (f(p)-f(-p))/\|\|f(p)-f(-p)\|\|</math>		Define <math>g:S^2\rightarrow S^1</math> by <math>g(p) = (f(p)-f(-p))/\|\|f(p)-f(-p)\|\|</math>

Trefor at 21:18, 8 February 2008

2008-02-08T21:18:17Z

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	Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>		Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>
	<math>(\gamma:I\rightarrow S^1, \gamma(0)=1, ind \gamma = \tilde{\gamma}(1))</math>		<math>(\gamma:I\rightarrow S^1, \gamma(0)=1, ind [\gamma] = \tilde{\gamma}(1))</math>

Trefor at 21:17, 8 February 2008

2008-02-08T21:17:40Z

← Older revision		Revision as of 17:17, 8 February 2008
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	Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>		Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>
	<math>(\gamma:I\rightarrow S, \gamma(0)=1, ind \gamma = \tilde{\gamma})</math>		<math>(\gamma:I\rightarrow S^1, \gamma(0)=1, ind \gamma = \tilde{\gamma}(1))</math>

Megan: /* First Hour */

2008-02-04T23:34:38Z

First Hour

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	Define a homotopy of paths <math>S^1\rightarrow S^1</math> by first setting <math>q(z) = p(z)/\|\|p(z)\|\|]\in S^1</math>		Define a homotopy of paths <math>S^1\rightarrow S^1</math> by first setting <math>q(z) = p(z)/\|\|p(z)\|\|]\in S^1</math>

	Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting on. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial <math>p_{\alpha}(z) = z^n + \alpha</math>(lower order terms) and let <math>\alpha</math> tend from 1 to 0 in the second half of the homotopy.		Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial <math>p_{\alpha}(z) = z^n + \alpha</math>(lower order terms) and let <math>\alpha</math> tend from 1 to 0 in the second half of the homotopy.

	Hence, at the end of the homotopy we are left with <math>z^n/\|\|z^n\|\|</math> on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction.		Hence, at the end of the homotopy we are left with <math>z^n/\|\|z^n\|\|</math> on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction.
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	'''Example:'''		'''Example:'''

	This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.		This is NOT true with 4 sections. For instance consider a triangular based pyramid inscribed inside a sphere with each face a different colour. Imagine a light bulb at the center of the sphere so that on the sphere we get 4 sections with 4 different colours from each side of the pyramid. Clearly none of them contains a pair of antipodes, yet are closed with a union of the whole sphere.


	===Second Hour===		===Second Hour===

Trefor: /* First Hour */

2008-01-18T23:27:02Z

First Hour

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Trefor at 05:13, 18 January 2008

2008-01-18T05:13:25Z

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==Class 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>

===First Hour===

We begin by reformulating our previous lemma into a more general form:

'''Lemma 1'''

Let <math>p:(X,x_0)\rightarrow(B,b_0)</math> be a covering map. Then every family of paths <math>\gamma:Y\times I\rightarrow B s</math>uch that <math>\forall y\in Y\ \gamma(y,0)=b_0</math> has a unique lift <math>\tilde{\gamma}:Y\times I\rightarrow X</math> such that

0) <math>\forall y \tilde{\gamma}(y,0) = x_0</math>

1) <math>\gamma = p\circ\tilde{\gamma}</math>

Claim 1: <math>Ind[\gamma]</math> is well defined, hence <math>\pi_1(S^1)\cong\mathbb{Z}</math>

<math>(\gamma:I\rightarrow S, \gamma(0)=1, ind \gamma = \tilde{\gamma})</math>

''Proof of Claim 1''

We construct the homotopy H between two such gamma's which, schematically, is a square with <math>\gamma_1</math> along the bottom, <math>\gamma_0</math> along the top and along the side the parameter Y where the homotopy is just horizontal lines between <math>\gamma_1</math> and <math>\gamma_0.</math>

Then, the lemma implies the existence of a homotopy <math>\tilde{H}</math> which is schematically a square with <math>\tilde{\gamma_1}</math> along the bottom, <math>\tilde{\gamma_0}</math> along the top, x=0 on the left and side and the right hand side taking values in <math>p^{-1}(1)</math>

''Proof of Lemma 1''

Recall the proof for the version of the lemma when we were dealing with a single path, not a family. We now just need to extend this proof to the case of having a family of paths. We know that for each <math>y\in Y</math> we can get a <math>\tilde{\gamma}</math>

We need to show that in fact the result is continuous. As continuity is a local property, we simply need to prove this in a neighborhood about <math>y_0</math>.

Now by a "good" open set in B, we mean that the preimage under p CAN be written as a product.

Hence, we choose a neighborhood about the first interval in <math>\mathbb{R}</math> extending from <math>y_0</math> (see proof of one path case for explanation) and this gets mapped to a "good" open set in B. As it is the image of an interval, it is compact in an open set, so can put a small neighborhood about the interval such that the image of the neighborhood is in the "good" open set in B.

Then, duplicating the proof of the earlier version of the lemma, this establishes continuity in the neighborhood.

'''Applications of <math>\pi_1(S^1)\cong\mathbb{Z}</math>'''

1) We get again the proof of non existence of a retract <math>r:D^2\rightarrow S^1</math>

Indeed, assume we DID have such a retract. We would have the following commuting diagram.

<math>\begin{matrix}
D^2&\rightarrow^{r}&S^1\\
\uparrow &\nearrow_{I}& \\
S^1&&\\
\end{matrix}</math>

Applying the functor <math>\pi_1</math> would yield the diagram

<math>\begin{matrix}
\{0\}&\rightarrow^{r}&\mathbb{Z}\\
\uparrow &\nearrow_{I}& \\
\mathbb{Z}&&\\
\end{matrix}</math>

But clearly this does not exist, as the only map from <math>\{0\}</math> is the trivial one.

Recall the non existence of such retracts implies ''Brouwer's Theorem''

2) '''Fundamental Theorem of Algebra'''

if <math>p\in\mathbb{C}[z]</math> is a polynomial with degree greater than zero then there <math>\exists z_0\in\mathbb{C}</math> such that <math>p(z_0) =0</math>

''Proof:''

Suppose not, i.e., there exists a polynomial p that has no roots. dividing by the coefficient of the highest order term, <math>p = z^n +</math> lower order terms.

Define a homotopy of paths <math>S^1\rightarrow S^1</math> by first setting <math>q(z) = p(z)/||p(z)||]\in S^1</math>

Now the homotopy starts with a base of q on the point 0. During the first half of the homotopy we grow the size of the loop on which q is acting on. At the halfway point, the loop is so large such that the first term in the polynomial dominates the lower order terms. Then in the second half of the homotopy, the lower order terms are shut off. More precisely we consider the polynomial <math>p_{\alpha}(z) = z^n + \alpha</math>(lower order terms) and let <math>\alpha</math> tend from 1 to 0 in the second half of the homotopy.

Hence, at the end of the homotopy we are left with <math>z^n/||z^n||</math> on a large loop. This is the generator we had previously, and is not the identity. Hence, we have a constant homotopic to something non trivial and this establishes the contradiction.

3) '''Boruk-Ulam'''

If <math>f:S^2\rightarrow \mathbb{R}^2</math> is continuous then <math>\exists p\in S^2</math> such that <math>f(p) = f(-p)</math>

Corollary:

1) <math>S^2</math> is not a subset of <math>\mathbb{R}^2</math> (cannot be embedded) as f is not 1:1

2) If <math>S^2 =A_1\cup A_2\cup A_3</math>, with <math>A_i</math> closed then at least one <math>A_i</math> contains a pair of antipodes.