In this
session all paths are in an open subset D of
C
and parametrized in [0,1].
Definitions.
Let
and
be
two paths in D.
1)
If
(0)=(0)=a,
say and
(1)=
(1)=b,
we say that
is
homotopic to
in
D
if there
is a continuous mapping
D such that
and
f
and
and
.
The
mapping is called a homotopy from
to
.
The above
identities define a family of paths
in D which connects
and
.
Intuitively, this means that
can be continuously deformed to
.
2) In
particular, closed paths
and
are
homotopic in D if there is a continuous
mapping
D such that
and
and
More
particularly, a closed path
is homotopic to a point a in
D if is homotopic to the path
defined by
.
Thus we
can now rephrase the definition of simple connected set
D to be a set in which every closed curve in D
is homotopic to a point in D.
Theorem.
Let
and let S denote the set of all
paths in D with initial point a and terminal
point b. Consider the binary relation H
defined on S by
H are
homotopic.
Then H
is an equivalence relation.
Proof.
1) H
is reflexive: If
S,
define
by
.
Since
is continuous, so is
. Moreover,
and
.
So
is a homotopy from
to
.
2)
H
is symmetric: Suppose
such that
H.
Let be a homotopy from
to
. Define
by
.
Clearly
is continuous. Moreover,
.
So
is a homotopy from
to
.
3)
H
is
transitive: Let
such that
and
. Let
and
be two homotopies from
to
and from
to
respectively. Define
by
if
and
if
. Note that
is continuous on
and [0,1]x( and at
we get
1)=. So
is continuous on (t,. Hence
is continuous on [0,1]x[0,1]. Moreover,
and
if
and
if
.
Similarly,
. Therefore
is a homotopy from
to
.
We now
begin to explore the impact of homotopy on integration
Definition.
Let
C
and let be a closed path in
C
-{a}. The index (or winding number) of
with respect to a, denoted by
, is defined by
.
We state
without proof some properties of the index in the
following
Theorem.
a) Let
be a closed path and
be its complement. Then
is an integer-valued function on
which is constant in each component of
and which is 0 in the unbounded
component of .
b) Let
and
be homotopic closed paths in
C
-{0}. Then
.
c)
Let and
be homotopic closed paths in
C
such that does not take the value 0 and such that
, then
does not take the value 0 and
.
Lemma.
If
and
are closed paths and if a is a
complex number such that
,
then
=.
Proof.
Note that
the inequality guarantees that
and
are in
C
-{a}. Let
. Then
and again using the inequality
. Thus the graph of
is contained in the disk
and hence outside
and therefore
by a) of the previous theorem.
Integrating with respect to t on [0,1]
gives the result.
As we’ll
see now, this lemma leads to a very important theorem
where if we add homotopy to the hypothesis of the lemma,
then we can dispose of the inequality (this shows the
impact of homotopy on the index integral and as we shall
see in later sessions on more general integrals):
Theorem.
Let
and
be homotopic closed paths in an open
subset D of
C.
Suppose . Then
=.
Proof.
Since and
be homotopic in D, there is a
continuous mapping
D such that
and
and
and
.
Since
is compact and the continuous image of
a compact set is compact,
is compact . Hence there exists
such that
. Moreover, by compactness,
is uniformly continuous. Thus there
exists a positive integer n such that
whenever
.
Define
polygonal closed paths
by
if
and
.
Thus
. In particular, taking the extreme
cases and
, we get
.Moreover,
.
On the
other hand, .
It
follows, from the last three inequalities and
applications of the previous lemma,
that a has the same index with respect to each of
the paths which completes the proof.
Example.
Let C
. Let be the positively oriented circle
such that
, then using the last theorem we have
=.
Thus , in
this case, =1 which explains the terminology of
winding number. |