I have following problem
Consider Poincare disk.
$i.e$ $\mathbb{M}=\{ (x,y) \in \mathbb{R}^2 : x^2+y^2 <1 \}$ with metric $ g= 4\frac{dx^2 +dy^2}{(1-x^2-y^2)^2}$
Show the complex mobius transformation, \begin{align} z \rightarrow \frac{z-w}{1-\bar{w}z} \end{align}understood as a map of $\mathbb{M}$ into itself.
Here one can set as $z=x+iy$, and $w=a+ib$, where $(x,y) \in \mathbb{M}$ and $(a,b) \in \mathbb{M}$.
I don't understand the sentence "a map of $\mathbb{M}$ into itself. " What does this mean?
I try to see $\frac{z-w}{1-\bar{w}z} \in \mathbb{M}$
\begin{align} z \rightarrow \frac{z-w}{1-\bar{w}z} \end{align} with $z=x+iy$, and $w=a+ib$, where $(x,y) \in \mathbb{M}$ and $(a,b) \in \mathbb{M}$.
\begin{align} \frac{z-w}{1-\bar{w}z}&=\frac{x+iy-(a+ib)}{1-(a-ib)(x+iy)}=\frac{x-a+i(y-b)}{1-ax-by + i(ay-bx)} \\ &=\frac{x-a+i(y-b)}{1-ax-by + i(ay-bx)} \frac{1-ax-by- i(ay-bx)}{1-ax-by- i(ay-bx)} \\ &= \frac{(x-a)(1-ax-by)+(y-b)(ay-bx)+i[(y-b)(1-ax-by)-(x-a)(ay-bx)]}{(a^2+b^2)x^2 + (a^2 +b^2)y^2 -2ax -2by +1} \\ &=\frac{ay^2 -ax^2 -2bxy +(a^2 +b^2 +1)x -a}{(a^2+b^2)x^2 + (a^2 +b^2)y^2 -2ax -2by +1} + i \frac{bx^2 +by^2 -2axy +(b^2 -a^2 +1)y-b}{(a^2+b^2)x^2 + (a^2 +b^2)y^2 -2ax -2by +1} \\ &=\tilde{a}+i\tilde{b} \end{align} Then how can i show $\tilde{a}, \tilde{b}$ is in $\mathbb{M}$?