Requirement for Connected Domain to be Simply Connected Domain

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Theorem

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


Then $D$ is simply connected if and only if 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.

$\blacksquare$


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