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.