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