# 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.

• 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. – Inactive - avoiding CoC Apr 19 '17 at 20:45
• @Servaes With your hint, I'm convinced that $x$ and $y$ belonging to different regions partitioned by $Q$ map to different regions partitioned by $f(Q)$, since by your proof, such $x$ and $y$ belong to $Q'$ which has a non-empty intersection with $Q$. Thus, at any point that $Q$ and $Q'$ intersect, we can define an angle between the tangent vectors obtained by moving along $Q$ and $Q'$. By conformality of $LFTs$, it follows that this angle is preserved in the image space, implying that $f(Q)$ and $f(Q')$ intersect. What if $x$ and $y$ belong to the same region? – Muno Apr 19 '17 at 20:56
• Applying a linear fraction transformation $f$ yields a line or circle $f(Q')$ containing $f(x)$ and $f(y)$ that intersects $f(Q)$ if and only if $x$ and $y$ are in the same region. This shows that $f(x)$ and $f(y)$ belong to the same region if and only if $x$ and $y$ do. – Inactive - avoiding CoC Apr 19 '17 at 21:48
• I fact, a linear fractional transformation ca be viewed as a rigid transformation of the Riemann sphere. See Mobius transformations revealed. – Mark McClure Apr 19 '17 at 22:42

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:

1. 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 $f(Q)$.

2. 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:

1. 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 $f(Q)$.

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