i might as well give a full solution. the proof for this is all over the internet and is in most textbooks that deal with functions indepth. it might as well be on here...or ami just re-inventing the wheel? ...or maybe bored.

Proof:

We proceed using a direct proof. Assume is invertible. Then is one-to-one and onto (there is a theorem that tells us this, it should be in your text).

We show is onto: Let . Since is onto, there exists such that . But that means with . Thus is onto.

We show is one-to-one: Assume with . Then , or in other words, . But is one-to-one, so . Hence, is one-to-one

QED