Let $G \subseteq \mathbb{C}$ be an open set, where $\mathbb{C}$ has the usual Euclidian topology.
Let $\gamma$ be a closed curve in $G$, i.e. a continuous map $\gamma : [0,1] \rightarrow G$ such that $\gamma(0) = \gamma(1)$. Write $\{ \gamma \}$ for its image.
Now, the set $\mathbb{C} - \{ \gamma \}$ is open, and exactly one of its connected components is unbounded.
Assume $\gamma$ is homotopic to zero in $G$. Show that $\delta G$ is contained in the unbounded component of $\mathbb{C} - \{ \gamma \}$.
Remarks:
For "homotopic to zero" I am following the convention of Conway's introductory book to complex analysis: There is a continuously varying family of curves deforming $\gamma$ into a constant curve, all the curves in this family are closed (i.e. same begin and endpoints) but the begin/endpoints of different curves in the family need not agree.
Attemps:
If you draw a picture it is immediately clear, yet I have failed to give a proper proof... We definitely need $\gamma$ homotopic to zero in $G$, for otherwise a counterexample is easily given ($\gamma$ any circle around $0$ and $G = \mathbb{C} - \{0\}$).
My idea would be to let $\Gamma : [0,1]^2 \rightarrow G$ be a homotopy deforming $\gamma$ into a constant path, and then show that the image of $\Gamma$ always contains all the bounded components of $\mathbb{C} - \{ \gamma \}$ (this would be interesting in itself!). Since a point of $\delta G$ cannot be contained in the image of $\Gamma$, this would show what we want to show.
