Theorem about inverse of an inverse f

[issacnewton]

Hi

I have to prove that if f:A\rightarrow B is bijective (onto and one to one)
then

f=(f^{-1})^{-1}

Following is my attempt. Since f:A\rightarrow B, we have f^{-1}:B\rightarrow A.

Now there is another theorem which says that function f is a bijection (onto and one to one)
iff inverse of f is also bijection. Let g:A\rightarrow B be the inverse of
f^{-1}. We know that g exists since f^{-1} is a one to one.
No let x \in A be arbitrary , then

\exists \,\, y \in B \backepsilon

g(x)=y but since

y \in B \Rightarrow \exists \,\, x_1 \in A \backepsilon

f(x_1)=y \Rightarrow f(x_1)=g(x)

So if I can show that x=x_1 and since x is arbitrary , I will complete the proof.
But I am stuck here. Can anybody suggest something ?

thanks
Newton

 

[hunt_mat]

Let g=(f^{-1})^{-1} and examine g\circ f^{-1} and f^{-1}\circ g what does this say about g?

 

[issacnewton]

Hi mat

Thanks for the hint. Here's my improvement. There is another theorem given in the section
in which I am doing this problem.

Theorem: Suppose that f:A\rightarrow B is a bijection, then

(a)(f^{-1}\circ f)(x) = x \;\;\forall x \in A

and

(b) (f \circ f^{-1})(y)=y \;\;\forall y \in B

So I think I can use this theorem here, since f^{-1} and g are inverses
of each other. and both of them are bijections since f itself is a bijection. We have that

g:A\rightarrow B \;\; \mbox{and}\;\; f^{-1}:B\rightarrow A

So using the stated theorem, we have

f^{-1}(g(x))=x \;\;\forall x \in A

\Rightarrow g(x) = f(x) \;\;\forall x \in A

In my last post , I got the equation f(x_1)=g(x). So from here can I claim that
x=x_1

If I can claim that, then I can say that

f(x)=(f^{-1})^{-1}(x) \;\;\forall x \in A

\Rightarrow f=(f^{-1})^{-1}

is it ok ?

 

[issacnewton]

mat , can you confirm my logic in the last post ?

 

[hunt_mat]

Looks good.

 

[issacnewton]

thanks for helping, mat