Mobius transformations and showing injections, surjections, and bijections Let $z_0$ be in the open upper half-plane $\mathbb{C}_{+}$. Show that the map $B_{z_0}(z) = \frac{z-z_0}{z-\overline{z_0}}$ is one-to-one from the closed upper half plane onto the closed unit disk. It is bijective from the real line onto the unit circle. And from $\mathbb{C}_{+}$ onto the open unit disk.

For injectivity, $z_1 = z_2 \implies f(z_1) = f(z_2)$. So we can show $\frac{z_2 - z_0}{z_2 - \overline{z_0}} = \frac{z_1 - z_0}{z_1 - \overline{z_0}}$ means $z_1 = z_2$ by algebra probably.
However, how do we relate the points to the closed upper half plane and the closed unit disk? And to the other spaces?
 A: I am still feeling complex analysis frisky.
As I'm too lazy to check, the injectivity of both arguments (I call it "both" because I'm thinking of this as two questions: show that $B_{z_{0}}$ maps $\mathbb{C}_+$ onto the open unit disk, and show that $B_{z_{0}}$ maps the real line bijectively onto the unit circle) is probably straightforward as you described, so I will not show either.
It should be pretty straightforward to show that $B_{z_0}(z)=\tfrac{z-z_0}{z-\bar{z}_0}$ is a map from the open upper half-plane onto the open unit disk as well. Indeed, let $w\in D$ ($D$ being the unit disk). Setting $w=\tfrac{z-z_0}{z-\bar{z}_0}$ and solving for $z$ (should) show
$$
z=\frac{w\bar{z}_0-z_0}{w-1}.
$$
(Here we need not worry that $z$ is undefined, because $w\in D$ implies that $w\neq 1$.) For $z$ to be in the upper half-plane we should have $\text{Im}(z)>0$. Since $\text{Im}(z)=\tfrac{1}{2i}(z-\bar{z})$ we compute
\begin{align*}
\text{Im}(z)&=\frac{1}{2i}\left(\frac{w\bar{z}_0-z_0}{w-1}-\frac{\bar{w}z_0-\bar{z}_0}{\bar{w}-1}\right)\\&=\frac{1}{2i}\left(\frac{\left(|w|^2\bar{z}_0-\bar{w}z_0-w\bar{z}_0+z_0\right)-\left(|w|^2z_0-\bar{w}z_0-\bar{z}_0w+\bar{z}_0\right)}{|w-1|^2}\right)\\
&=\frac{1}{2i}\frac{|w|^2(\bar{z}_0-z_0)+(z_0-\bar{z}_0)}{|w-1|^2}\\
&=\frac{\text{Im}(z_0)-|w|^2\text{Im}(z_0)}{|w-1|^2}\\
&=\frac{(1-|w|^2)\text{Im}(z_0)}{|w-1|^2}.
\end{align*}
Since $w\in D$, $(1-|w|^2)>0$, and since $z_0\in\mathbb{C}_+$, $\text{Im}(z_0)>0$. Thus $\text{Im}(z)>0$ and $z\in\mathbb{C}_+$. This shows that $B_{z_0}(z)$ in a bijective map from the open upper half-plane to the open unit disk. Through similar means you should be able to show that the $B_{z_0}$ maps the real line to the unit circle bijectively. These two facts together show that $B_{z_0}$ maps the closed upper half-plane to the closed unit disk.
