Show that the path homotopy definition of simple connectedness is equivalent to the loop homotopy definition of connectedness (in the Complex Plane).
That is, every two paths in a subset of the complex plane with the same endpoints are homotopic if and only every loop in that subset complex plane is homotopic to the trivial loop.
Suppose that every two paths with in the same endpoints in some $D \subset \mathbb{C}$ were homotopic ($D$ is simple connected is the path homotopy sense).
Let $\gamma$ be any closed curve in $D$. Let $x_{1}$ and $x_{2}$ be any two different points on $\gamma$. This breaks $\gamma$ into two parts, let them be $\gamma_{1}$ and $\gamma_{2}$. These are curves with the same endpoints ($x_{1}$ and $x_{2}$). They must be homotopic because they are contained in $D$. Let $\Gamma$ be their homotopy. So $\Gamma(0,t) = \gamma_{1}(t)$ and $\Gamma(1,t) = \gamma_{2}(t)$.
A homotopy between $\gamma$ and the trivial loop at $x_{1}$ can be obtained by "pushing" $\gamma_{1}$ to $\gamma_{2}$ together and then shrinking down $\gamma_{2}$ to $x_{1}$. This looks like this
$$\Phi(s,t) = \begin{cases} \Gamma (s,t) \qquad 0 \leq s, t \leq 1 \\ \Gamma (1, (2-s)t) \qquad 1 \leq s \leq 2 \, \text{ and } \, 0 \leq t \leq 1 \\ \Gamma (1, (2-s)(2-t)) \qquad 1 \leq s, t \leq 2 \end{cases} $$
I think this covers the half the problem. But I'm not really sure how to go the other way.