Conformal map between $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ and $\{z \in \mathbb{C} \mid 0 < \operatorname{Im}(z) < 7\}$ As it says in the title, I am looking for a conformal map from $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\{z \in \mathbb{C} \mid 0 < \operatorname{Im}(z) < 7\}$, but with the following restriction on the boundary components: $(-\infty, -1]$ is mapped to $\operatorname{Im}(z) = 7$ and $[1, \infty)$ is mapped to $\operatorname{Im}(z) = 0$.
So far I have been able to map $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\{w \in \mathbb{C} \mid 0 < \operatorname{Im}(w) < 7\}$, but I don't know if the boundary components are mapped to their counterparts in the desired way. I used the following sequence of maps (note, the branch of the logarithm is always the one with argument $(0, 2\pi)$):


*

*$z \mapsto\frac{1}{\sqrt{2}}\sqrt{z - 1}$ maps $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\mathbb{H}\setminus\{bi \mid b \in [1, \infty)\}$.


*$z \mapsto \dfrac{z-i}{z+i}$ maps $\mathbb{H}\setminus\{bi \mid b \in [1, \infty)\}$ to $\mathbb{D}\setminus[0, 1)$.


*$z \mapsto \sqrt{z}$ maps $\mathbb{D}\setminus[0, 1)$ to the upper half disc.


*$z \mapsto z + \frac{1}{z}$ maps the upper half disc to the lower half plane.


*$z \mapsto \log z$ maps the lower half plane to $\{z \in \mathbb{C} \mid \pi < \operatorname{Im}(z) < 2\pi\}$.


*$z \mapsto \frac{7}{\pi}(z - \pi i)$ maps $\{z \in \mathbb{C} \mid \pi < \operatorname{Im}(z) < 2\pi\}$ to $\{z \in \mathbb{C} \mid 0 < \operatorname{Im}(z) < 7\}$.

The after applying the first and second map, $(-\infty, -1]$ is mapped to $[0, 1)$. What happens to this boundary component under the third map? It seems like it should remain unchanged, but geometrically it seems that it should be mapped to $(-1, 1)$.
Is there an alternative approach to this problem which would make it easier to see what happens to the boundary components?
Note, if we could construct a map as desired, except it swapped the boundary components, we could define a new map by (post)composing with the map $z \mapsto 7 - z$; this new map would then have all the desired properties.
 A: As Mike Miller said, boundary slits should be thought of as "two-sided curves", and conformal maps are apt to split the two sides: formally, they may have different boundary limits depending on the direction of approach to the same boundary point. The simplest example of $z\mapsto \sqrt{z}$ mapping $\mathbb C\setminus (-\infty,0]$ onto half-plane: two sides of the negative semi-axis go in two different places. 
For your problem, I  would start with 
$$z\mapsto \frac{1+z}{1-z}$$
which transforms the domain into $\mathbb C\setminus (-\infty,0]$. The two boundary pieces are separated by point $-1$ (it came from $\infty$, which separated them in the original domain).
Then square root onto right halfplane; the two pieces are separated by $\pm i$: one is between $\pm i$ and one is outside. Use a fractional linear transformation to send them to $0$ and $\infty$: 
$$z\mapsto \frac{z+i}{z-i}$$ 
This was a preparation for applying the logarithm. The logarithm breaks the boundary of halfplane at $0$, sending two parts onto two sides of infinite strip. Now only cosmetics remain: moving the strip where you want it to be.
A: The sine function maps the strip $\{z\in \mathbb{C}\mid |\operatorname{Re}(z)|<\tfrac{\pi}{2}\}$ biholomorphically onto $\mathbb{C}\setminus \{x\in \mathbb{R}\mid |x|\geq 1\}$. So start with $z\mapsto\arcsin(z)$ and compose with a suitable linear transformation.
