# Thread: Every bijection has a two sided inverse

1. ## Every bijection has a two sided inverse

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.

2. Originally Posted by oldguynewstudent
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.
You want to show that $\displaystyle f$ is a bijection.

Suppose that $\displaystyle f(a)=f(b)$ then you know that
$\displaystyle f^{-1}(f(a))=a~\&~f^{-1}(f(b))=b$.
Because $\displaystyle f(a)=f(b)$ that means $\displaystyle a=b$.

You seem to have done the surjective part.