null-homotopic $\Leftrightarrow$ free-null-homotopic

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! :)

Thanks!

• 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 – Maxime Ramzi Jul 18 '19 at 16:28
• @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 $x_0$. – Paul Frost Jul 19 '19 at 16:51
• @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 – 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.