Let $f:A\to B$ be a function. Then show that f  is a bijection if and only if there exists a function $g:B\to A$ such that $fog=I_B$ and $gof=I_A$ and in this case $g=f^{-1}$

Given $f:A\to B$ be a function

Case(i): If f is bijection then prove $fog=I_B$ and $gof=I_A$ in this case $g=f^{-1}$ 

Since $f:A\to B$ is a bijection 

$\Rightarrow f^{-1}:B\to A$ is also a bijection and we know that $f^{-1}\mathit{\text{of}}=I_A,\mathit{\text{fof}}{}^{-1}=I_B$ 

Now take $g=f^{-1}$

$\Rightarrow gof=I_A$ and $fog=I_B$

Case (ii) :

Conversely assume that there exists a function 

$g:B\to A$ such that $gof=I_A$and $fog=I_B$

Then we have to prove f  is bijection and $g=f^{-1}$ 

$Let$ $a_1,a_2\in A$

$\Rightarrow f\left(a_1\right),f\left(a_2\right)\in B$ $(\because f:A\to B$ is a function$)$

Let $f\left(a_1\right)=f\left(a_2\right)$ 

$\Rightarrow g\left(f\left(a_1\right)=g\left(f\left(a_2\right)\right.\right.$ $(\therefore g:B\to A$ is a function $)$

$\Rightarrow \left(gof\right)\left(a_1\right)=\left(\mathit{\text{gof}}\right)\left(a_2\right)\Rightarrow I_A\left(a_1\right)=I_A\left(a_2\right)$

$\Rightarrow a_1=a_2$ 

$\therefore f$ is one-one 

Let $b\in B$

$\Rightarrow g\left(b\right)=a\in A(\because g:B\to A$ is a function $)$

$\therefore f\left(a\right)=f\left(g\right(b\left)\right)=\left(fog\right)\left(b\right)=I_B\left(b\right)=b$

$\therefore f$ is onto 

$\therefore f$ is both one-one $\&$onto

$\Rightarrow f:A\to B$ is bijection and is invertible function so $f^{-1}:B\to A$ is bijective

$\therefore g\text{ and }{f}^{-1}$ have same domain B

Let $b\in B$ since $f:A\to B$ is onto, then there exists $a\in A$ such that $f\left(a\right)=b$ 

$\Rightarrow f^{-1}\left(b\right)=a$

Now $g\left(b\right)=g\left(f\right(a\left)\right)$

                 $=\left(gof\right)\left(a\right)=I_A\left(a\right)$

                $=\left(gof\right)\left(a\right)=I_A\left(a\right)$

Hence $g=f^{-1}$