Homotopy

Definition:

Given topological spaces $\QTR{Large}{S}$ and $\QTR{Large}{T}$ and continuous maps MATH and MATH, we say that $\QTR{Large}{f}$ is homotopic to $\QTR{Large}{g}$ if there exists a continuous map MATH such that for all MATH, MATH and MATH

If $\QTR{Large}{f}$ is homotopic to $\QTR{Large}{g}$ ,we use the notation $\QTR{Large}{f} $ MATH $\QTR{Large}{g}$

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Examples:

Example 1 Let MATH be the natural inclusion map, and let MATH be the constant map MATH . Let

MATHbe defined by MATH.

Example 2 Let MATH MATHbe the circle defined by the parametric equation ,

MATH
graphics/homotopy__24.png

and let MATH be the ellipse MATH .
graphics/homotopy__27.png

Let

MATHbe defined by MATH

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Theorem:

Homotopy is an equivalence relation.

Proof:

  1. $\QTR{Large}{f}$ MATH $\QTR{Large}{f}$: By the homotopy MATH

  2. $\QTR{Large}{f}$ MATH MATH $\QTR{Large}{f}$: If MATH is a homotopy from $\QTR{Large}{f}$ to $\QTR{Large}{g}$ then MATH is a homotopy from $\QTR{Large}{g}$ to $\QTR{Large}{f.}$

  3. $\QTR{Large}{f}$ MATH $\QTR{Large}{g}$ and $\QTR{Large}{g}$ MATH MATH $\QTR{Large}{h.}$

Given MATH a homotopy from $\QTR{Large}{f}$ to $\QTR{Large}{g}$ and MATH a homotopy from $\QTR{Large}{g}$ to $\QTR{Large}{h}$ we define a function

MATH for MATH

and

MATH for MATH

Note that this is well defined since MATH

What is left to show is that the function is continuous. To do this we invoke the push out construction.

Let $\QTR{Large}{S}_{1}$ and $\QTR{Large}{S}_{2}$ be two topological space homeomorphic to $\QTR{Large}{S\ }$but disjoint from $\QTR{Large}{S\ }$and each other. Consider the push out diagram

$\vspace{1pt}$

MATH

MATH

MATH

defined by MATHfor MATH

and MATHfor MATH

We are implicitly using the various homeomorphisms between $\QTR{Large}{S}_{1}$ , $\QTR{Large}{S}_{2}$, and $\QTR{Large}{S}_{1}$

Consider the diagram:

MATH

MATH

MATH

We have at once that $\QTR{Large}{H}$ is continuous since

MATHfor MATH

and

MATHfor MATH

What still needs to be verified is that the identity map is a homeomorphism between MATH with the push out topology and MATH with the product topology.

To see this we look at the diagram

MATH

MATH

MATH

where the right most MATH is considered to have the product topology. Again,

Homework Due Thursday April 27

verify that MATH is continuous hence $\QTR{Large}{1}$ is continuous.

and,

under $\QTR{Large}{1}$ ,the image of closed sets are closed hence the inverse of $\QTR{Large}{1}$ is continuous with respect to the two topologies.