Mobius transformations are bijections proof

Question

I don't understand the last line of this proof. To show a function is bijective we need to show it is one-to-one and onto. The proof shows that $f$ is one-to-one only. For some reason $f^{-1}$ being one-to-one shows that $f$ is onto, I guess?

Answer 1

I won't provide a formal proof, but in an intuitive sense since a function maps all elements to an element of the codomain, and the inverse function is the mapping of those elements to their preimage, if the inverse function is a function and is one-to-one, then every element in the codomain is mapped to every element of the domain by the inverse function, and every element in the domain is mapped to every element of the codomain. So the function is onto as well.