I have read in quite a few books the statement that "A mobius transformation gives a bijection from the extended complex plane to itself." How do I prove this? I know that you can find a unique Mobius transformation sending three points to three points, am I then looking for a transformation that sends three points to $0,1,\infty$ or is there a more specific way to prove this?
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It seems as though you've actually shown that the group of Möbius transformations acts transitively on triples of points in the extended complex plane. Namely, if $(a,b,c)$ and $(x,y,z)$ are two triples of points in the extended complex plane then you've proven that there exist Möbius transformations $A,B$ (which I'll regard as matrices) such that $A(a,b,c)=(0,1,\infty)$ and $B(x,y,z)=(0,1,\infty)$. As cmk pointed out in his comment, this means that $B^{-1}A$ will map $(a,b,c)$ to $(x,y,z)$. Moreover, $A^{-1}B$ will map $(x,y,z)$ to $(a,b,c)$. If you restrict to the first coordinate then you have a map from the extended complex plane to itself, together with its inverse. This gives you your desired bijection.
As noted above by cmk, say $$ f(z) = \frac{az+b}{cz+d}. $$ Note that $$ f^{-1}(w)=\frac{dw-b}{-cw+a}. $$
Using this inverse you can easily prove that for any $w$ in the extended complex plane there is a $z$ s.t. $f(z)=w$.