I believe I have the same professor G as mathgirl13 and the whole class is struggling.

I need to prove the converse of the above:

If f: X $\displaystyle \longrightarrow $ Y has a two-sided inverse, prove that it is a bijection.

Given $\displaystyle (f^{-1}\circ f)(x) = x $ and $\displaystyle (f^{-1}\circ f)(y) = y$

T.P. $\displaystyle (f\circ f^{-1})(y1) = (f\circ f^{-1})(y2) \longrightarrow y1 = y2$

$\displaystyle (f\circ f^{-1})(y1) = y1 = (f\circ f^{-1})(y2) = y2 $

Now choose $\displaystyle (f^{-1})(y) =x$. This should prove f is injective, right?

Next T.P. $\displaystyle \forall y \epsilon Y \exists x \epsilon X | f(x) = y$

Choose y=f(x) then $\displaystyle (f^{-1})(f(x)) =y $

but also $\displaystyle (f^{-1})(f(x)) = f(x) therefore f(x)=y \forall y \epsilon Y$

This should prove f is surjective, right?

Thanks for any and all help.