# Prove Inverse of a bijective function is also bijective

1. Jan 21, 2013

### blindgibson27

1. The problem statement, all variables and given/known data

Prove that the inverse of a bijective function is also bijective.

2. Relevant equations

One to One
$f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}$

Onto
$\forall y \in Y \exists x \in X \mid f:X \Rightarrow Y$
$y = f(x)$

3. The attempt at a solution
It is to proof that the inverse is a one-to-one correspondence. I think I get what you are saying though about it looking as a definition rather than a proof.

Let $f:X\rightarrow Y$ be a one to one correspondence, show $f^{-1}:Y\rightarrow X$ is a one to one correspondence.

$\exists x_{1},x_{2} \in X \mid f(x_{1}) = f(x_{2}) \Leftrightarrow x_{1}=x_{2}$

furthermore, $f^{-1}(f(x_{1})) = f^{-1}(f(x_{2})) \Rightarrow f^{-1}(x_{1}) = f^{-1}(x_{1})$ (by definition of function $f$ and one to one)

kind of stumped from this point on..
I may want to transfer this post over to the hw section though, I did post to just get a confirmation on my thoughts on bijection but it is now turning into something a bit more specific than that

2. Jan 21, 2013

### jbunniii

To show that $f^{-1}$ is an injection, we need to show that $f^{-1}(x_1) = f^{-1}(x_2)$ implies $x_1 = x_2$. What happens if you apply $f$ to both sides of $f^{-1}(x_1) = f^{-1}(x_2)$?

3. Jan 21, 2013

### blindgibson27

My thoughts are that $f^{-1}$ cancels by definition,

so to begin let $f^{-1}(x_{1}) = f^{-1}(x_{2})$
$\Leftrightarrow f(f^{-1}(x_{1})) = f(f^{-1}(x_{2}))$ since $\forall y \in Y$ the function $f^{-1}(y) = x$ for some $x \in X$
$\Leftrightarrow x_{1} = x_{2}$

Last edited: Jan 21, 2013
4. Jan 21, 2013

### HallsofIvy

Staff Emeritus
Yes, that is correct and shows that, because f is injective, so is $f^{-1}$.

To show that $f^{-1}$ is surjective, you want to prove that, for any x in X, there exist y in Y such that $$x= f^{-1}(y)$$.

Given any x in X, let y= f(x). Then use the other direction: $f^{-1}(y)= f^{-1}(f(x))= x$.

5. Jan 21, 2013

### blindgibson27

many thanks HallsofIvy