I have some troubles with a problem, I`ve been dealing for a while, so i hope you can help me. :)

I need to show, that, if $G \subset \mathbb{C}$ is a Domain and $ \gamma : [0,1] \rightarrow G$ is a closed path, then $\gamma$ is FEP-null-homotopic if and only if it is free-null-homotopic.

First I will explain the phrases, as we defined it.

Two paths are (FEP-)homotopic, if there exists a continous map $h: [0,1] \times [0,1] \rightarrow G$, such that both of the paths have the same beginning- and the same end-Point. A closed path is called (FEP-)null-homotopic, if it is in the same equivalence-class (formed by the homotopy-relation) as it's start point. This means, that we can shrink the path to it's start-point with this map.

Two closed paths $\gamma, \eta$ are free-homotopic, if the continous map $h: [0,1] \times [0,1] \rightarrow G$ holds $h(0,\tau) = h(1, \tau)$ and $h(.,0)=\gamma$ and $h(.,1)=\eta$

Now after giving all these definitions, I'll explain my thoughts (please correct me, if I'm misstaking):

FEP-null-homotopic means, that the paths can shrink to a point, wheareas the starting (=end)point is fixed. Free-homotopic means, that this point is allowed to be translated in the $\mathbb{C}$-plane.

So I suppose the "$\Rightarrow$" - direction should be finished, due to we have the special-case of a free homotopy without translation.

For the "$\Leftarrow$"-direction I think I need to show, that for each pair of free-homotopic paths there exists a continous map between them, which is homotopic. This should be possible via "moving along" with the corresponding point. The movement/translation of this point is continous, because the free-homotopic map $h$ is continous, so it is also continous for fixed $t$ in the second argument (I think this describes the "movement" of my fixed point.)

So I suppose it should be possible to create such a map, which "deletes" the movement, such that we get the wanted homotopic map.

Well, this has been my idea for a possible solution. My Problem is, that I don't know how to make a feasible proof out of that (providing, that my idea is right) and how to define the claimed homotopy.

I hope some of you can help me! :)


| cite | improve this question | | | | |
  • $\begingroup$ Do you know about fundamental groups ? If so, there is a proof that is valid for any space : if $\gamma, \eta$ are freely homotopic, then they are conjugate in the fundamental group; so if one is trivial, so is the other $\endgroup$ – Maxime Ramzi Jul 18 '19 at 16:28
  • $\begingroup$ @Max To apply this, you have to invoke the (admittedly fairly obvious) fact that if a loop based at $x_0$ is freely null-homotopic (i.e. freely homotopic to some constant loop based at any $x_1$) is also freely homotopic to the constant loop based at $x_0$. $\endgroup$ – Paul Frost Jul 19 '19 at 16:51
  • $\begingroup$ @PaulFrost : of course, but if the OP doesn't know about fundamental groups anyway, then the comment won't evolve into much more than just a comment, so I didn't get into that much detail $\endgroup$ – Maxime Ramzi Jul 19 '19 at 17:18

As you observed, $\Rightarrow$ is obvious.

To show $\Leftarrow$, we can proceed as follows.

Let $D^2 = \{ z \in \mathbb C \mid \lvert z \rvert \le 1\}$. Define map $p : [0,1] \times [0,1] \to D^2, p(t,s) = (1-s)e^{2\pi i t}$. This is a continuous surjection. It maps $[0,1] \times \{s \}$ onto the circle with radius $1-s$ and center $0$ (draw a picture). Now let $h$ be a free homotopy from $\gamma$ to a constant loop $c$. Consider $(t,s), (t',s') \in [0,1] \times [0,1]$ such that $p(t,s) = p(t',s')$. This implies $1-s = \lvert p(t,s) \rvert = \lvert p(t',s') \rvert = 1-s$, i.e. $s = s' = \sigma$. For $\sigma < 1$ we conclude $e^{2 \pi i t} = e^{2 \pi i t'}$. Hence $t = t'$ or $t, t' \in \{0,1\}$. Therefore $h(t,s) = h(t',s')$. For $\sigma = 1$ we get $h(t,s) = h(t,0) = c(t) = c(t') = h(t',s')$. Therefore again $h(t,s) = h(t',s')$.

This shows that $h': D^2 \to G, h'(z) = h(t,s)$ with any $(t,s) \in p^{-1}(z)$ is well-defined. We have $h' \circ p = h$. This implies that $h'$ is continuous since $p$ is a quotient map. (Here we invoke well-known facts from general topology. A direct proof for the continuity of $h'$ will be given later.)

Now define a map $q : [0,1] \times [0,1] \to D^2, q(t,s) = s + (1-s)e^{2\pi it}$. This maps $[0,1] \times \{s \}$ onto the circle with radius $1-s$ and center $s$. We have $q(t,0) = p(t,0), q(t,1) = 1, q(0,s) = q(1,s) = 1$ for all $t,s$.

Define $H = h' \circ q : [0,1] \times [0,1] \to G$. This a continuous homotopy such that

(1) $H(t,0) = h'(q(t,0)) = h'(p(t,0)) = h(t,0) = \gamma(t)$

(2) $H(i,s) = h'(q(i,s)) = h'(1) = h'(p(i,0)) = h(i,0) = \gamma(i) = a$ for $i = 0,1$

(3) $H(t,1) = h'(q(t,1)) = h'(1) = a$

This shows that $\gamma$ is null-homotopic.

Let us finally verify that $h'$ is continuous. So Let $A \subset G$ be closed. Then $h^{-1}(A)$ is closed in $[0,1] \times [0,1]$, hence compact. Thus $p(h^{-1}(A))$ is compact, hence closed in $D^2$. But $(h')^{-1}(A) = p(p^{-1}((h')^{-1}(A))) = p((h' \circ p)^{-1}(A)) = p(h^{-1}(A)$ which shows that $(h')^{-1}(A)$ is closed.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.