Prove Inverse of a bijective function is also bijective

[blindgibson27]

Homework Statement

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

Homework 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)

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.

How about this..

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 homework 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

 

[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)?

 

[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}

 

[HallsofIvy]

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.

 

[blindgibson27]

many thanks HallsofIvy