0708-1300/Class notes for Tuesday, January 15

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Class 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.

First Hour

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

Lemma 1

Let $p:(X,x_{0})\rightarrow (B,b_{0})$ be a covering map. Then every family of paths $\gamma :Y\times I\rightarrow Bs$uch that $\forall y\in Y\ \gamma (y,0)=b_{0}$ has a unique lift ${\tilde {\gamma }}:Y\times I\rightarrow X$ such that

0) $\forall y{\tilde {\gamma }}(y,0)=x_{0}$

1) $\gamma =p\circ {\tilde {\gamma }}$

Claim 1: $Ind[\gamma ]$ is well defined, hence $\pi _{1}(S^{1})\cong \mathbb {Z}$

$(\gamma :I\rightarrow S,\gamma (0)=1,ind\gamma ={\tilde {\gamma }})$

Proof of Claim 1

We construct the homotopy H between two such gamma's which, schematically, is a square with $\gamma _{1}$ along the bottom, $\gamma _{0}$ along the top and along the side the parameter Y where the homotopy is just horizontal lines between $\gamma _{1}$ and $\gamma _{0}.$

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

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 $y\in Y$ we can get a ${\tilde {\gamma }}$

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 $y_{0}$.

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 $\mathbb {R}$ extending from $y_{0}$ (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 $\pi _{1}(S^{1})\cong \mathbb {Z}$

1) We get again the proof of non existence of a retract $r:D^{2}\rightarrow S^{1}$

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

${\begin{matrix}D^{2}&\rightarrow ^{r}&S^{1}\\\uparrow &\nearrow _{I}&\\S^{1}&&\\\end{matrix}}$

Applying the functor $\pi _{1}$ would yield the diagram

${\begin{matrix}\{0\}&\rightarrow ^{r}&\mathbb {Z} \\\uparrow &\nearrow _{I}&\\\mathbb {Z} &&\\\end{matrix}}$

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

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

2) Fundamental Theorem of Algebra

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

Proof:

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

Define a homotopy of paths $S^{1}\rightarrow S^{1}$ by first setting $q(z)=p(z)/||p(z)||]\in S^{1}$

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 $p_{\alpha }(z)=z^{n}+\alpha$(lower order terms) and let $\alpha$ tend from 1 to 0 in the second half of the homotopy.

Hence, at the end of the homotopy we are left with $z^{n}/||z^{n}||$ 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 $f:S^{2}\rightarrow \mathbb {R} ^{2}$ is continuous then $\exists p\in S^{2}$ such that $f(p)=f(-p)$

Corollary:

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

2) If $S^{2}=A_{1}\cup A_{2}\cup A_{3}$, with $A_{i}$ closed then at least one $A_{i}$ contains a pair of antipodes.