# degree of a path, path homotopy.

• Mar 18th 2009, 03:08 PM
math8
degree of a path, path homotopy.
Let f and g be loops in the unit circle S^1 with base point (1,0). We show if deg(f)=deg(g), then f and g are path homotopic.

I know f,g: [0,1]-->S^1 and there must be unique paths f* and g* (lifting paths) s.t. f*(0)=0 and g*(0)=0 and s.t. pof*=f, pog*=g where p:R-->s^1 is defined by p(t)=(cos 2 pi t, sin 2 pi t).

I also know that deg(f)= f*(1), deg(g)=g*(1) which are integers.

I was thinking about coming up with a path homotopy F:[0,1]x[0,1]-->S^1 s.t. for each t in [0,1]; F(t,0)=f(t) and F(t,1)=g(t)
for each s in [0,1]; F(0,s)=(1,0) and F(1,s)=(1,0).

But I am not sure what path homotopy would work here.
• Mar 18th 2009, 07:44 PM
aliceinwonderland
The Covering Homotopy Property: If $\displaystyle F: I \times I \rightarrow S^1$ is a homotopy such that F(0,0) = 1, then there is a lifting of F to a unique covering homotopy $\displaystyle \bar{F}: I \times I \rightarrow R$ for which $\displaystyle \bar{F}(0,0)=0$.

You have loops f and g in $\displaystyle S^1$ with a base point 1 having the same degree. This implies that the covering paths $\displaystyle \bar{f}, \bar{g}$ beginning at common initial point 0 have common terminal point $\displaystyle \bar{f}(1)=\bar{g}(1)$. The following homotopy $\displaystyle H:I \times I \rightarrow R$ demonstrates the equivalence of $\displaystyle \bar{f},\bar{g}$.

$\displaystyle H(t,s) = (1-s)\bar{f}(t) + s\bar{g}(t), (t,s) \in I \times I$.

By the Covering Homotopy Property, the homotopy pH is a path homotopy between f and g (p is defined in your question).

You need to make sure the following cases if pH is indeed a path homotopy between f and g.

pH(0,s), pH(1,s), pH(t,0), and pH(t,1).
• Mar 19th 2009, 06:56 AM
math8
Thanks, this makes a lot of sense. But I am wondering, the Covering Homotopy Property says if http://www.mathhelpforum.com/math-he...905671ee-1.gif is a homotopy such that
F(0,0) = (1,0), then there is a lifting of F to a unique covering homotopy
F*:IxI -->R for which F*(0,0)=0 and poF* = F

But does this imply that if H* is a path homotopy between f* and g* s.t. H*(0,0)= 0, then poH* must be a path homotopy between f and g? (where f* and g* are the liftings of f and g respectively).

Because it seems that the proof is based on that fact.
• Mar 19th 2009, 04:39 PM
aliceinwonderland
Quote:

Originally Posted by math8
Thanks, this makes a lot of sense. But I am wondering, the Covering Homotopy Property says if http://www.mathhelpforum.com/math-he...905671ee-1.gif is a homotopy such that
F(0,0) = (1,0), then there is a lifting of F to a unique covering homotopy
F*:IxI -->R for which F*(0,0)=0 and poF* = F

But does this imply that if H* is a path homotopy between f* and g* s.t. H*(0,0)= 0, then poH* must be a path homotopy between f and g? (where f* and g* are the liftings of f and g respectively).

Because it seems that the proof is based on that fact.

To define H shown above, f and g should be loops in $\displaystyle S^1$ with the base point (1,0) having the same degree. Otherwise, the covering paths $\displaystyle \bar{f}, \bar{g}$ do not have the common terminal point.

By defintion,

$\displaystyle deg(f) = \bar{f}(1)$ and $\displaystyle deg(g) = \bar{g}(1)$.

So the degree of f and g should be the same to define H shown above.

The Covering Homotopy Property simply ensures the existence of a unique covering homotopy $\displaystyle \bar{F}$ of F. If f and g are equivalent loops in $\displaystyle S^1$, the degree of f and g are the same and we can define H shown above.

$\displaystyle deg(f)=\bar{f}(1)=\bar{F}(1,0)=\bar{F}(1,1)=\bar{g }(1)=deg(g)$. (F is defined in your post.)

$\displaystyle \bar{F}$ and H are basically the same for the above case. The latter is intended to represent the homotopy starting off with $\displaystyle \bar{f}$ and $\displaystyle \bar{g}$.

Now, we can conclude that

For loops f and g in S^1 with base point (1,0), [f]=[g] iff deg(f) = deg(g).
• Mar 19th 2009, 05:37 PM
math8
Thanks, this really helps.