Uniqueness of conformal mappings with different normalizations: three boundary points, or an interior point Some stuff I've seen in lecture but am still a little shaky on:
1)  To determine my mapping explicitly, it suffices to know where 3 distinct points on my pre-image object, say, the unit circle, gets mapped to.  Say I construct my mapping so that 3 points, 1,0,-1 goes to 1,0,-1, respectively.  Why is this enough?  I've done some problems, using the cross-ratio and solving for my mapping, w(z), but I don't know with confidence why mapping 3 points is enough.  
2)  Sticking with the example above, how come knowing where just one point on the interior of the pre-image circle gets mapped to...tells us where all of the interior of the pre-image circle gets mapped to - in this case, it would be either the interior of the image circle or the exterior of the image circle.  I vaguely know that this comes from a connectedness argument from introductory real analysis (advanced calculus); that a continuous mapping maps connected sets to connected sets - as well as compact sets to compact sets.  I feel like ...if most of the interior of the pre-image disk gets mapped to the interior of the image disk, but some of the pre-image points in the disk gets mapped to ...the exterior of the image disk, isn't the image...still connected / path-connected?
I know that LFTs must map circles and lines to circles or lines.  Is this the key point that I am not considering? That perhaps...the boundary, i.e., the circle "separates" the image plane into two disjoint sets, so if preimage points land on the exterior of the image disk, then, in fact, my image is not path-connected?
 A: You mostly answered your own questions already, but because I wanted to think through this myself, too, I decided to write something.
1: Consider a Möbius transformation (a linear fractional transformation with the $ad - bc \neq 0$ property) $\;f: \bar{\mathbb{C}} \to \bar{\mathbb{C}}$. Choose three distinct points $z_1, z_2, z_3 \in \bar{\mathbb{C}}$. Then then there exists a unique circle or a line that goes through the points $z_1, z_2, z_3$. This is the reason why it's enough to know where 3 distinct points map to. The points $z_1, z_2, z_3$ define a unique circle/line in the domain and the points $f(z_1), f(z_2), f(z_3)$ define a unique circle/line in the codomain. Now we know that the circle/line that goes through the points $z_i$ maps to a circle/line that goes through the points $f(z_i)$, so we know the whole image of the whole circle/line in the domain.
2: Now consider an image of some disk $f \mathbb{D}(z_0,r)$. We can find the image of $\partial \mathbb{D}(z_0,r)$ by choosing three points from it and looking where they map to. Then the set $S := f\partial \mathbb{D}(z_0,r)$ divides the codomain to two "pieces". If $S$ is a circle, we have the inside and the outside of the circle.
We also know that $f\mathbb{D}(z_0,r) \cap S = \emptyset$, because $\mathbb{D}(z_0,r) \cap \partial \mathbb{D}(z_0,r) = \emptyset$ and $f$ is an injection (because Möbius transformations are bijections). Assume that some point $z \in \mathbb{D}(z_0,r)$ maps to the inside. Now if there exists a point $z' \in \mathbb{D}(z_0,r)$ that maps to the other side, the set $f\mathbb{D}(z_0,r)$ would have to somehow "jump" over the set $S$, because it can't overlap with it. This can't happen because $f$ is continuous.
This idea can be made rigorous easily with connectivity. Choose an open disk $D$ s.t. $\partial D = S$. Now if $f \mathbb{D}(z_0,r) \cap D \neq \emptyset$ and $\operatorname{ext}D \cap f \mathbb{D}(z_0,r) \neq \emptyset$, where $\operatorname{ext}D$ is the exterior of $D$, from topology we know that $f \mathbb{D}(z_0,r) \cap S$ must be non-empty too (because $f\mathbb{D}(z_0,r)$ is connected), which is impossible by a previous argument.
Then there's the question why we know that $f\mathbb{D}(z_0,r)$ fills the whole of $D$, but this follows from the fact that $\operatorname{ext} \mathbb{D}(z_0,r)$ can't have any points that map to $D$, so $f \mathbb{D}(z_0,r)$ must fill whole of $D$ because of bijectivity.
