Requirement for Connected Domain to be Simply Connected Domain

Theorem
Let $D \subseteq \C$ be a connected domain.

Then $D$ is simply connected all paths in $D$ with the same initial points and end points are path-homotopic.

That is, given two paths in $D$:


 * $\gamma: \closedint a b \to D$
 * $\sigma: \closedint c d \to D$

with $\map \gamma a = \map \sigma c$ and $\map \gamma b = \map \sigma d$, there exists a continuous function $H: \closedint 0 1 \times \closedint 0 1 \to D$ such that:


 * $\map H {s, 0} = \map \gamma s$ for all $s \in \closedint 0 1$
 * $\map H {s, 1} = \map \sigma s$ for all $s \in \closedint 0 1$
 * $\map H {t, 0} = \map \gamma t$ for all $t \in \closedint 0 1$
 * $\map H {t, 1} = \map \sigma t$ for all $t \in \closedint 0 1$

Proof
From Open Domain is Connected iff it is Path-Connected, it follows that $D$ is path-connected.

The results now follows by the definition of simply connected sets.