# Homotopies between circles and paths in the annulus {z: 0<a<|z|<b }

1. Mar 3, 2009

### WWGD

Hi, everyone :
I read a problem posted somewhere else on showing that, "for the correct value m--
an integer" , any curve lying on an annulus A={0<a<|z|< b} was homotopic to the
circle re^(i*pi*m*t).

So I thought of this question: what are the homotopy classes of the annulus?.
I know that Pi_1(A)=Z , since the annulus deformation-retracts to S^1 , so that
the two are homotopic.

I wonder if a homotopy H:X-->Y ,in general tells us how to map homotopy
classes in X, to those in Y; specifically:

We have that loops that go around the circle n times form a class, i.e., loops
that go around once are not homotopic to those going around more than once, etc.

So, knowing this , and knowing that S^1 is homotopic to the annulus: can we
figure out what the homotopy classes are in the annulus?

Thanks .

2. Mar 3, 2009

### quasar987

Recall that two spaces X and Y that are of the same homotopy type have isomorphic fundamental groups because if f:X-->Y is such a homotopy equivalence, then the induced map f_* is an isomorphism. In the event that X is a deformation retracts of Y, then the inclusion of X into Y is a homotopy equivalence and so allows you to populate the homotopy classes of Y quite easily!

Is this what you were asking about at all?

3. Mar 6, 2009

### wofsy

I find your question hard to understand. What do you mean by homotopy classes?

If any two curves in the annulus are homotopic to a curve that winds around a circle within it as you seem to say in the first paragraph, what else do you need to know?

4. Mar 6, 2009

### WWGD

Well, to try to rephrase it more clearly, my question is this : ( I think Quasar987 answered
it, If I understood him well ), in a more general way :

Assume both X,Y are path-connected spaces.
Say we know that X,Y are homotopically equivalent spaces, and that we know Pi_1(X),
but we also know the "structure" of Pi_1(X)=G (no need for basepoint , by assumption of path-connectedness.), in the sense that we know the actual isomorphism between the loops
in X and G.
How can we use above information to find out the structure of Pi_1(Y)?.

Now, to the specifics of our case :

In our case, we have X=S^1 , and Pi_1(X)=Z, and Y=A, the annulus , and we know
that the "structure" of Pi_1(X) is given by :

h: Pi_1(X) --->Z given by :

loop that goes around n times ---> n in Z

(with n<0 if loops go in opposite direction)

Can we use this to figure out the structure of Pi_1(A)?. I mean, can we know which
paths in the annulus A correspond to the same element n in Pi_1(S^1)?.
Can we also figure out with this information which paths in Pi_1(S^1) correspond
to the same element n in Pi_1(A) ?

Unlike the case of the circle, a lot wilder things can happen with paths in A:
non-smoothness, self-intersection, etc., and I at least, cannot easily tell which
paths in A are equivalent , in the sense of representing the same element n in
Pi_1(A)=Z .

5. Mar 7, 2009

### WWGD

Thanks, Quasar987. I know my posts can be unclear. I agree with you; I am just having
trouble working with this idea in the specific case of S^1 and A, i.e., I am having trouble
making sense of f_* ([n]) in A , for some reason, where [n] is the class of paths that go
around S^1 n times. Problem is that f in this case, if I did get things correct, is just the
inclusion map , since the homotopy equiv. between S^1 and A is given by:

f: S^1-->A is inclusion
g:A --->S^1

is the deformation of the inner ring to the outer ring, or viceversa. Then f_* ([n])=

i_*([n]) =([i o n]) , and I cannot make much sense of this.

6. Mar 7, 2009

### wofsy

Your i_*([n]) =([i o n]) just says that the homotopy class in the annulus of a path that is made by first choosing a path that winds around the circle n times then mapping this path into the annulus via the inclusion map is the image under the induced homomorphism of groups of the homotopy class of the path in the circle that winds n times around.

i is a mapping of topological spaces, i_* is a mapping of groups.

An arbitrary path in the fundamental group of the circle can be very wild just as in the annulus. It can be non-differentiable, self-intersecting and what not. In fact it can be non-differentiable at every point. There is no simplicity to the closed paths in the circle.

What you are really after is an isomorphism of the integers with the fundamental group of the circle. Once you have this the annulus follows immediately from the retraction map.

7. Mar 7, 2009

### WWGD

What you are really after is an isomorphism of the integers with the fundamental group of the circle. Once you have this the annulus follows immediately from the retraction map.[/QUOTE]

I thought I had that (the isp between Pi_1(S^1) and Z ) already, with [n]-->n ;
[n] being the class of paths in S^1 that go around once, and what I needed now
was the isomorphism between Pi_1(A) and Z, which , as we talked, can be
obtained by using the homotopy equiv. (def. retract) between S^1 and A.

8. Mar 8, 2009

### wofsy

I thought I had that (the isp between Pi_1(S^1) and Z ) already, with [n]-->n ;
[n] being the class of paths in S^1 that go around once, and what I needed now
was the isomorphism between Pi_1(A) and Z, which , as we talked, can be
obtained by using the homotopy equiv. (def. retract) between S^1 and A.[/QUOTE]

right. I just though because you said that the paths in the annulus were wild compared to the circle that you might not have realized that the circle can also have wild paths. After all in the annulus all you need to do is draw a circle and take the paths that wind around it. This gives you its fundamental group. Also after retraction, a wild path in the annulus could still be wild. It will not generally be a simple wind around.

9. Mar 9, 2009

### quasar987

Like you said, the inclusion i:S^1-->A induces an isomorphism $$i_*:\pi_1(\mathbb{S}^1)\rightarrow \pi_1(A): [\gamma]\mapsto [i\circ \gamma]$$ between the fundamental groups. And you know that $$\pi_1(\mathbb{S}^1)$$ is actually made up only of $$[e^{2\pi imt}]$$ for m integer. Hence $$\pi_1(A)$$ is only made up of $$[e^{2\pi imt}]$$ for m integer also.

So modulo that radius r, this shows that for the correct value the integer m, any curve lying on the annulus A={0<a<|z|< b} is homotopic to the circle re^(i*2pi*m*t).