Originally Posted by

**matt.qmar** hello!

I am trying to work this out, not sure where to start even though the theorem appears intuitive! I thought I had proven it but then realised I has assumed the existance of an inverse!

For functions f,g and sets A, B

Given $\displaystyle f: A \rightarrow B $and $\displaystyle g: B \rightarrow A $ and $\displaystyle g \circ f = 1_A, f \circ g = 1_B $

where $\displaystyle 1_A$ is the function $\displaystyle 1_A(a) = a, \forall a \in A $

Then $\displaystyle f^{-1} $ exists and $\displaystyle g = f^{-1} $

To show the exsistance of an inverse requires that f is a bijection? not sure how to derive that from what is given.

Thanks in advance!