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!
 A: 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.
