Lifting Homotopies

We will need the following:

Lemma

Let $\QTR{Large}{S}$ be any function of to a topological space $\QTR{Large}{S}$ . Suppose we are given two sequences of reals

and then $\QTR{Large}{h}$ is continuous if and only if for any $\QTR{Large}{i<m}$ and $\QTR{Large}{j<n}$ the restriction is continuous. We denote this restriction as

Proof:

That is continuous, given $\QTR{Large}{h}$ is continuous, is by definition.

The other direction is just a special case of the push out lemma. Given the are continuous, note that we are partitioning as a finite union of closed subspaces so for any closed $\QTR{Large}{S}$ we have MATH , a finite union of closed sets.

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An Explicit Covering:

We will work with the covering of by two open sets.

Since .

MATH

and

MATH

and are homeomorphisms with explicit inverses

and

We are identifying $\QTR{Large}{(x,y)}$ pairs with suitable "continuous" ranges of angles

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Theorem:

The winding number is invariant under relative homotopy. that is if are relatively homotopic then

Proof:

In detail, the hypothesis states that we are given a map $\QTR{Large}{S^{1}}$ such that . That is for all $\QTR{Large}{t\in }$ $\QTR{Large}{I.}$

Moreover and

We will show that we can find a map such that for all $\QTR{Large}{t\in }$ $\QTR{Large}{I}$ and for all

This will prove the theorem since and hence

Thus, since $\QTR{Large}{I}$ is connected is constant. So

To show that the homotopy $\QTR{Large}{F}$ can be "lifted" to a map $\QTR{Large}{G}$ having the given properties.

Consider the covering . MATH is a covering of , hence has a Lesbegue number . Choose any and such that Thus for each $\QTR{Large}{i<m}$ and $\QTR{Large}{j<n}$ or (or both).

We now define $\QTR{Large}{G}$ inductively using the sequence

We are given for all $\QTR{Large}{t\in }$ $\QTR{Large}{I}$ .

Suppose we have defined for all $\QTR{Large}{t\in }$ $\QTR{Large}{I}$ .

Fix $\QTR{Large}{j<n\ }$ and for notational simplicity assume that .

Since is connected . Define on

What is left to check is that these maps agree on common edges. But, on common edges, since they start at the same value and cover the same map, this is just requires a restatement of the unique path lifting lemma of last lecture.