Complement of the image of a loop in the plane Let $w_{1}, w_{2} \in \mathbb{D}$ be such that $\Im\left( w_{1} \right) < 0$ and $\Im\left( w_{2} \right) > 0$, where $\mathbb{D}$ denotes the open unit disk around  $0$ in $\mathbb{C}$. Suppose $\gamma^{-}$ is a path joining $w_{1}$ to $w_{2}$ in $\mathbb{D} \setminus \mathbb{R}_{-}$ and $\gamma^{+}$ is a path joining $w_{2}$ to $w_{1}$ in $\mathbb{D} \setminus \mathbb{R}_{+}$, and set $\Gamma = \gamma^{-}\left( [0, 1] \right) \cup \gamma^{+}\left( [0, 1] \right)$.
How can one prove -- with elementary tools if possible -- that the connected component of $\mathbb{C} \setminus \Gamma$ containing $0$ is contained in $\mathbb{D}$?
 A: Here's the standard "winding number" proof. If this doesn't quite seem elementary to you, you can read up on closed and exact forms, with an emphasis on $d\theta$ in $\mathbb C - \{0\}$. Here is a wikipedia link with some details.
Orient the path $\gamma_-$ from $w_1$ to $w_2$, and orient $\gamma_+$ from $w_2$ to $w_1$. Let $\gamma$ be the concatenated path $\gamma = \gamma_- * \gamma_+$, which is a closed curve based at $w_1$ and may be regarded as parameterization of the "loop" $\Gamma$.
On $\mathbb D - \mathbb R_-$ let $\theta_- \in (-\pi,+\pi)$ be the angle coordinate, and on $\mathbb D - \mathbb R_+$ let $\theta_+ \in (0,2\pi)$ be the angle coordinate. Notice that that on the upper half plane we have $\theta_+ = \theta_-  \in (0,\pi)$, and on the lower half plane we have $\theta_+ = \theta_- + 2\pi \in (-\pi,0)$, and in either case we have $d\theta_+ = d\theta_-$ which I'll denote as just $d\theta$, and you can use the formula
$$d\theta = \frac{x \, dy - y \, dx}{x^2+y^2}
$$
I'm following the traditional abuse of notation here: $d\theta$ stands for a closed 1-form, but it's not an exact form, i.e. it is not actually equal to "$d$" of any function defined on the entirety of its domain which is $\mathbb C - \{0\}$. 
Now let's integrate, using the fundamental theorem of calculus for line integrals:
\begin{align}\int_\gamma d\theta &= \int_{\gamma_-} d\theta_- + \int_{\gamma_+} d\theta_+ \\
&= (\theta_-(w_2)-\theta_-(w_1)) + (\theta_+(w_1) - \theta_+(w_2)) \\
&= (\theta_-(w_2) - \theta_+(w_2)) + (\theta_+(w_1) - \theta_-(w_1)) \\
&= 0 + 2\pi \\
&= 2\pi
\end{align}
So, now let $U$ be the component of $\mathbb C \setminus \Gamma$ containing $0$.
Arguing by contradiction, let's suppose that $U$ is not contained in $\mathbb D$. 
Connected open subsets of $\mathbb C$ are path connected, so I can construct a continuous function $r :[0,\infty) \mapsto U$ such that $r(0)=0$, $r[0,1) \subset \mathbb D$, $r(1) \in \partial \mathbb D$, and $r[1,\infty)$ is a ray shooting out to infinity. Furthermore, by carefully perturbing $r$ I can guarantee that $r$ is one-to-one. Think of $r$ as a ray that is forced to be somewhat squiggly inside $\mathbb D$ in order to get past $\Gamma$, but once it manages to do that it shoots straight out to infinity just like any other respectable ray. But despite its early squiggly nature, it follows that $\mathbb C - \text{image}(r)$ is simply connected. 
Since $\mathbb C - \text{image}(r)$ is contained in the domain $\mathbb C - \{0\}$ of $d\theta$, and since $d\theta$ is an closed form, it follows that $d\theta$ is exact on the simply connected set $\mathbb C-\text{image}(r)$, i.e. there is an actual function $\theta_r : \mathbb C - \text{image}(r) \to \mathbb R$ such that $d\theta_r = d\theta$ on $\mathbb C - \text{image}(r)$. It follows that
$$\int_\gamma d\theta = \int_\gamma(d\theta_r) = \theta_r(w_1) - \theta_r(w_1)=0
$$
which gives the desired contradiction.
A: My answer is a rewording of Lee Mosher's answer. Nevertheless, I still post it since I find it more elementary -- it only uses basic complex analysis.
Suppose that the paths $\gamma^{-}$ and $\gamma^{+}$ are piecewise continuously differentiable and denote by $\gamma = \gamma^{-} \ast \gamma^{+}$ the concatenation of $\gamma^{-}$ and $\gamma^{+}$ -- that is, $$\gamma \colon t \in [0, 1] \mapsto \begin{cases} \gamma^{-}(2 t) & \text{if } t \in \left[ 0, \frac{1}{2} \right]\\ \gamma^{+}(2 t -1) & \text{if } t \in \left[ \frac{1}{2}, 1 \right] \end{cases} \, \text{.}$$ Let $\arg_{-} \colon \mathbb{C} \setminus \mathbb{R}_{-} \rightarrow (-\pi, \pi)$ (respectively $\arg_{+} \colon \mathbb{C} \setminus \mathbb{R}_{+} \rightarrow (0, 2 \pi)$) be the "principal value" of the argument on $\mathbb{C} \setminus \mathbb{R}_{-}$ (respectively $\mathbb{C} \setminus \mathbb{R}_{+}$). Then $z \in \mathbb{C} \setminus \mathbb{R}_{-} \mapsto \log \lvert z \rvert +i \arg_{-}(z)$ (respectively $z \in \mathbb{C} \setminus \mathbb{R}_{+} \mapsto \log\lvert z \rvert +i \arg_{+}(z)$) is a primitive of the map $z \mapsto \frac{1}{z}$ on $\mathbb{C} \setminus \mathbb{R}_{-}$ (respectively $\mathbb{C} \setminus \mathbb{R}_{+}$). Therefore, we have $$ \begin{split} \text{Ind}_{\gamma}(0) &= \frac{1}{2 i \pi} \left( \int_{\gamma^{-}} \frac{1}{z} \, dz +\int_{\gamma^{+}} \frac{1}{z} \, dz \right)\\ &= \frac{1}{2 \pi} \left( \arg_{-}\left( w_{2} \right) -\arg_{-}\left( w_{1} \right) +\arg_{+}\left( w_{1} \right) -\arg_{+}\left( w_{2} \right) \right)\\ &= 1 \end{split}$$ since $\arg_{+}\left( w_{1} \right) = \arg_{-}\left( w_{1} \right) +2 \pi$ and $\arg_{+}\left( w_{2} \right) = \arg_{-}\left( w_{2} \right)$, where $\text{Ind}_{\gamma}(0)$ denotes the index of $0$ with respect to $\gamma$. It follows that $\text{Ind}_{\gamma} \equiv 1$ on the connected component of $\mathbb{C} \setminus \Gamma$ containing $0$. Now, note that, since $\Gamma \subset \mathbb{D}$ and $\mathbb{C} \setminus \mathbb{D}$ is unbounded and connected, $\mathbb{C} \setminus \mathbb{D}$ is contained in the unique unbounded connected component of $\mathbb{C} \setminus \Gamma$, and hence $\text{Ind}_{\gamma} \equiv 0$ on $\mathbb{C} \setminus \mathbb{D}$. Thus, the connected component of $\mathbb{C} \setminus \Gamma$ containing $0$ is contained in $\mathbb{D}$.
