Given topological spaces
and
and continuous maps
and
,
we say that
is homotopic to
if there exists a continuous map
such that for all
,
and
If
is homotopic to
,we use the notation
Example 1 Let
be the natural inclusion map, and let
be the constant map
. Let
be
defined by
.
Example 2 Let
be
the circle defined by the parametric equation ,
and let
be the ellipse
.
Let
be
defined by
Homotopy is an equivalence relation.
:
By the homotopy
:
If
is
a homotopy from
to
then
is
a homotopy from
to
and
Given
a
homotopy from
to
and
a
homotopy from
to
we define a function
for
and
for
Note that this is well defined since
What is left to show is that the function is continuous. To do this we invoke the push out construction.
Let
and
be
two topological space homeomorphic to
but
disjoint from
and
each other. Consider the push out diagram
defined by
for
and
for
We are implicitly using the various homeomorphisms between
,
,
and
Consider the diagram:
We have at once that
is continuous since
for
and
for
What still needs to be verified is that the identity map is a homeomorphism
between
with
the push out topology and
with
the product topology.
To see this we look at the diagram
where the right most
is considered to have the product topology. Again,
Homework Due Thursday April 27
verify that
is continuous hence
is continuous.
and,
under
,the
image of closed sets are closed hence the inverse of
is continuous with respect to the two topologies.