The Winding Number

Definition and Notation

The Setting

This is the picture:

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Lemma:

Let MATH and let $\QTR{Large}{y\in }$ $\QTR{Large}{S^{1}}$. MATH

such that.

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Definition:

Suppose we are given a map MATH such that MATH and such that MATH then, MATH is called the winding number or degree of $\QTR{Large}{g.}$

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Lemma:

Let MATH such that MATH and such that MATHthen MATH

Proof:

Let MATH for allMATH such that MATHLet MATH. By the usual limit argument MATH . We want to show that $\QTR{Large}{l=1}$.

Let MATH . Since $\QTR{Large}{e}$ is single-valued (in fact a homeomorphism) on MATH, for MATH

But MATH and MATH thus if MATH then there is

some MATH MATHwith MATH

We want to show that MATH But follows from the fact that MATH and MATH

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Lemma:

Let MATH MATH then there exists a unique MATH , such that MATH and such that $\QTR{Large}{eg=f}$ . In particular we can define a winding number on $\QTR{Large}{[I,}$ MATH by setting

MATH

Proof:

Start with MATH and, using the fact that $\QTR{Large}{e}$ is a local homeomorphism define $\QTR{Large}{g}$ on the "largest possible" subinterval of $\QTR{Large}{I}$ .Now use a "lub" argument to show you can define it on all of $\QTR{Large}{I}$.

Homework due May 2. Give a detailed proof of the lemma:

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Theorem:

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

Proof(to follow)

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Consequences

  1. The maps MATH MATH and the constant map MATH MATH , where MATH are not relatively homotopic.

    Proof: MATH and MATH

  2. There does not exist a map (always read continuous) MATH MATHsuch that MATHis the identity map.

    Proof: $\QTR{Large}{e}$ and $\QTR{Large}{c}$ considered as maps in MATH MATH are relatively homotopic. The formula is

    MATH

    Here is the picture

    composing this relative homotopy with $\QTR{Large}{r}$ would give a relative homotopy of $\QTR{Large}{e}$ and $\QTR{Large}{c}$ in $\QTR{Large}{[I,}$ MATH , contradicting 1.

  3. Every map MATH MATHsuch that MATHis the identity map, is onto.

    Proof: If not we could find a map contradicting 2.

  4. Every map MATH MATHsuch that MATHis the identity map, has a fixed point.

    Proof: If not we could find a map contradicting 2.

Homework due May 2. prove 3. and 4. (hint the proofs are almost identical to the MATH case.