We write for a pair consisting of a topological space and subspace .
We write for a map such that
We write for the set of such maps.
Given we say that is relatively homotopic to if there is a homotopy from to such that . It is immediate that relatively homotopy is also an equivalence relation.
We will be concerned with the Real Line the Circle , which we take to be and
We will also be concerned with the (continuous) map defined by the formula
The reason that we use the parameter "" rather than "" will become clear in the in the next lecture.
We will be concerned with maps such that Using the notation introduced above, .
Since there will be no ambiguity we use the notation as an abreviation for .
Finally, we will be concerned with maps such that and such that In particular,
This is the picture:
Let and let .
such that.
(in fact we could set )
is a homeomorphism of open subintervals.
Suppose we are given a map such that and such that then, is called the winding number or degree of
Let such that and such that then
Let for all such that Let . By the usual limit argument . We want to show that .
Let . Since is single-valued (in fact a homeomorphism) on , for
But and thus if then there is
some with
We want to show that But follows from the fact that and
Let then there exists a unique , such that and such that . In particular we can define a winding number on by setting
Start with and, using the fact that is a local homeomorphism define on the "largest possible" subinterval of .Now use a "lub" argument to show you can define it on all of .
Homework due May 2. Give a detailed proof of the lemma:
The winding number is invariant under relative homotopy. that is if are relatively homotopic then
The maps and the constant map , where are not relatively homotopic.
Proof: and
There does not exist a map (always read continuous) such that is the identity map.
Proof: and considered as maps in are relatively homotopic. The formula is
Here is the picture
composing this relative homotopy with would give a relative homotopy of and in , contradicting 1.
Every map such that is the identity map, is onto.
Proof: If not we could find a map contradicting 2.
Every map such that is 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 case.