The images of disjoint regions separated by a circle or line are disjoint under a linear fractional transformation It is known that linear fractional transformations (LFTs) take lines and circles to either a line or a circle. 
$$ T: \{ \text{ Line or Circle } \} \mapsto \{ \text{ Line or Circle } \}$$
I would like to know whether the two regions of $\mathbb{C}$ partitioned by a line or circle map to regions that are disjoint with respect to each other. Additionally, are the images of these regions still separated by the image of the divider (the line or the circle)?
In multivariate real analysis, it is known that given a set $S$, a diffeomorphism $f$ takes
$$int(S) \mapsto int(f(S))$$ $$bd(S) \mapsto bd(f(S))$$ $$int(S^{c}) \mapsto int(f(S)^{c})$$ 
Perhaps there is an analog for holomorphic functions -- in particular, for LFTs.
 A: Here's an approach with a classical geometric flavour. First note the following:

Let $Q$ be a line or circle partitioning $\Bbb{C}$ into two regions and let $x,y\in\Bbb{C}-Q$. Then there exists a line or circle $Q'$ with $x,y\in Q$ and $Q'\cap Q=\varnothing$ if and only if $x$ and $y$ are in the same region. 

Now let $Q$ be a line or circle and let $f$ be a linear fraction transformation. We distinguish two cases:


*

*Suppose $x$ and $y$ are in the same region with respect to $Q$. Let $Q'$ be a line or
circle with $x,y\in Q'$ and $Q'\cap Q=\varnothing$. Then $f(Q')$ is a line or circle with $f(x),f(y)\in f(Q')$ and $f(Q')\cap f(Q)=\varnothing$, so
$f(x)$ and $f(y)$ are in the same region with respect to $f(Q)$.

*Suppose $x$ and $y$ are not in the same region with respect to $Q$. If $f(x)$ and $f(y)$ are in the same region with respect to $f(Q)$, then there exists a line or circle $Q'$ such that $f(x),f(y)\in Q'$ and $Q'\cap f(Q)=\varnothing$. Then $f^{-1}(Q')$ is a line or circle with $x,y\in f^{-1}(Q')$ and $f^{-1}(Q')\cap Q=\varnothing$, contradicting the fact that $x$ and $y$ are not in the same region with respect to $Q$. So $f(x)$ and $f(y)$ are in the same region with respect to $f(Q)$.

Here's an approach with a more topological flavour. First note the following:

Let $X$ be a connected topological space and $f: X\ \longrightarrow\ Y$ a continuous map. Then $f(X)$ is a connected subspace of $Y$

Now let $Q$ be a line or circle and let $f$ be a linear fraction transformation. Then the connected components of $\Bbb{C}-Q$ are the two regions that $Q$ divides $\Bbb{C}$ into. We distinguish two cases:


*

*Suppose $x$ and $y$ are in the same region with respect to $Q$. The restriction of $f$ to this region is a continuous map from a connected space to $\Bbb{C}-f(Q)$, so its image is connected. Ihis means it is contained in a connected component of $\Bbb{C}-f(Q)$, i.e. $f(x) $ and $f(y)$ are in the same region with respect to $f(Q)$.

*Suppose $x$ and $y$ are not in the same region. If $f(x)$ and $f(y)$ are in the same region with respect to $f(Q)$...
