- #1

issacnewton

- 1,026

- 36

I have to prove that if [itex]f:A\rightarrow B[/itex] is bijective (onto and one to one)

then

[tex]f=(f^{-1})^{-1}[/tex]

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

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 [itex]g:A\rightarrow B[/itex] be the inverse of

[itex]f^{-1}[/itex]. We know that g exists since [itex]f^{-1}[/itex] is a one to one.

No let [itex] x \in A [/itex] be arbitrary , then

[tex] \exists \,\, y \in B \backepsilon [/tex]

[tex] g(x)=y [/tex] but since

[tex] y \in B \Rightarrow \exists \,\, x_1 \in A \backepsilon [/tex]

[tex] f(x_1)=y \Rightarrow f(x_1)=g(x) [/tex]

So if I can show that [itex]x=x_1[/itex] and since x is arbitrary , I will complete the proof.

But I am stuck here. Can anybody suggest something ?

thanks

Newton