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 $\gamma :Y\times I\rightarrow B$ such 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 $\pi _{1}(S^{1})\cong \mathbb {Z}$ $(\gamma :I\rightarrow S^{1},\gamma (0)=1,ind[\gamma ]={\tilde {\gamma }}(1))$

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 $\exists z_{0}\in \mathbb {C}$ 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. 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.

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.

Second Hour

Proof of Corollary:

Consider $f:S^{2}\rightarrow \mathbb {R} ^{2}$ by $p\mapsto$ (dist to $A_{1}$, dist to $A_{2}$)

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 $A_{i}$.

Definition:

1) $\gamma :S^{1}\rightarrow S^{1}$ is even if $\gamma (-x)=\gamma (x)\ \forall x$

2) $\gamma$ is odd if $\gamma (-x)=-\gamma (x)\ \forall x$

Lemma:

1) If $\gamma$ is even then ind $\gamma$ is even

2) If $\gamma$ is odd then ind $\gamma$ is odd

Proof of Borsuk-Ulam:

Assume $f:S^{2}\rightarrow \mathbb {R} ^{2}$ has no p with f(p)=f(-p)

Define $g:S^{2}\rightarrow S^{1}$ by $g(p)=(f(p)-f(-p))/||f(p)-f(-p)||$

$\gamma =g|_{equator}:S^{1}\rightarrow S^{1}$ is odd. Therefore, $[\gamma ]\in \pi _{1}(S^{1})$ is non zero.

But $\gamma$ is homotopic to zero, a contradiction (the homotopy is a series of circles of decreasing radius where a point on the equator is fixed and its antipodal point is moved in an arc towards it)

Proof of part 1 of lemma:

Suppose $\gamma :S^{1}\rightarrow S^{1}$ is even

We can make a commuting diagram where $\gamma :S^{1}\rightarrow S^{1}$ is the same as going from $S^{1}\rightarrow S^{1}$ first via $z\mapsto z^{2}$ and then from $S^{1}\rightarrow S^{1}$ via $\lambda (z)=\gamma ({\sqrt {z}})$

We now consider this diagram under the functor $\pi _{1}$

We get $\mathbb {Z} \rightarrow ^{\times 2}\mathbb {Z} \rightarrow ^{\lambda _{*}}\mathbb {Z}$

which commutes with $\mathbb {Z} \rightarrow ^{\gamma _{*}}\mathbb {Z}$ along the bottom

$[\gamma ]=\gamma _{*}[\gamma _{1}]=\lambda _{*}(2[\gamma _{1}])=2\lambda _{*}([\gamma _{1}])$ which is even.

(Note, by $\gamma _{1}:S^{1}\rightarrow S^{1}$ we mean the identity)

Unfortunately, there is no simple adaption of this proof to deal with the odd case and so we introduce some more machinery first.

Definition

A topological group is a topological space G with a group structure such that all group operations are continuous

Examples:

1) $S^{1}$ (with rotation giving the group structure)

2) SO(3) (See a previous lecture for more info)

If G is a topological group there are two "products" we can define on $\pi _{1}(G)$:

1) $[\gamma ][\lambda ]=[\gamma \circ \lambda ]$ with $\circ$ being simply concatenation

2) $[\gamma ]*[\lambda ]=[\gamma *\lambda ]$ where $(\gamma *\lambda )(t)=\gamma (t)\cdot \lambda (t)$

Claim:

These two are the same and, in fact, are commutative. I will not attempt to describe the homotopy schematic here.

We return to the proof of the odd case of the lemma:

Assume $\gamma$ is odd. Then $\gamma *\gamma _{1}$ is even so $2\mathbb {Z} \in [\gamma *\gamma _{1}]=[\gamma \circ \gamma _{1}]=[\gamma ]+[\gamma _{1}]=[\gamma ]+1$. Therefore, $[\gamma ]$ is odd.

Note: The addition above is done in $\pi _{1}(S^{1})$ which is just $\mathbb {Z}$

Q.E.D.

Claim:

If $f,g:X\rightarrow Y$ are homotopic then $f_{*}=g_{*}:\pi _{1}(X)\rightarrow \pi _{1}(Y)$

Consider a loop $\gamma$ in X. Consider the images of this in Y under f and g. They look like two loops with the same base point and the homotopy H collapses them down to the same loop.

Then, $f\circ \gamma$ is homotopic to $g\circ \gamma$ using $H\circ \gamma$ as the homotopy.

Theorem

1) $\sim$ is an equivalence relation.

2) It is an ideal in the category of topological spaces. That is, $f\sim g\Rightarrow f\circ h\sim g\circ h$ and $h\circ f\sim h\circ g$ for h's such that these make sense.

Definition

X and Y are "homotopy equivalent" if they are isomorphic in {Topological Spaces} / {homotopy of maps}

I.e., $\exists f,g$ so that $f\circ g\sim I_{Y}$ and $g\circ f\sim I_{X}$ for $f:X\rightarrow Y$ and $g:Y\rightarrow X$

Example

1) $\mathbb {R} ^{n}$ is homotopy equivalent to a point via f which takes $\mathbb {R} ^{n}$ to a point and g is the zero map

2) An annulus is homotopy equivalent to $S^{1}$. Indeed, consider the map f which collapses the annulus to the circle in its middle, and g is just the identity.

3) Likewise for the figure "thick A" which is a subset of the plane in the shape of an A with everything being thick. There is a circle around the hole in the A. The map f just takes A to this circle and g is the identity.

4) The mobius band is also equivalent to a circle where f just collapses to the boundary circle.

This last example gives us some idea of the limitation, or "lack of subtlety" to our invariants given that we would all agree there is something fundamentally different between the mobius band, the annulus and the circle, homotopy equivalence is not sensitive to this difference.

Claim

If X and Y are homotopically equivalent then $\pi _{1}(X)\cong \pi _{1}(Y)$

Proof:

Consider $X\rightarrow ^{f}Y$ and $Y\rightarrow ^{g}X$ forming a commuting diagram and lets consider its image under $\pi _{1}$

Then we get $\pi _{1}(X)\rightarrow ^{f_{*}}\pi _{1}(Y)$ and $\pi _{1}(Y)\rightarrow ^{g_{*}}\pi _{1}(X$) as a commuting diagram.

We know that $f\circ g$ is homotopic to the identity, and thus we also get $f_{*}\circ g_{*}$ homotopic to the identity

Q.E.D.

Theorem

$\pi _{1}(X\times Y)\cong \pi _{1}(X)\times \pi _{1}(Y)$

Example: $\pi _{1}(\mathbb {T} ^{2})=\pi _{1}(S^{1}\times S^{1})=\mathbb {Z} \times \mathbb {Z} =\mathbb {Z} ^{2}$