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.