Lifting Homotopies

We will need the following:

Lemma

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

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

Proof:

That MATH 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 MATH are continuous, note that we are partitioning MATH as a finite union of closed subspaces so for any closed MATH $\QTR{Large}{S}$ we have MATH , a finite union of closed sets.

__________________________________________________________________

An Explicit Covering:

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

Since MATH.

MATH

and

MATH

MATH and MATH are homeomorphisms with explicit inverses

MATH

and

MATH

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

__________________________________________________________________

Theorem:

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

Proof:

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

Moreover MATHand MATH

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

This will prove the theorem since MATH and hence MATH

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

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

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

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

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

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

Since MATH is connected MATH . Define MATH on MATH

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.