Let $f:A\to B,g:B\to C$ be bijections. Then show that $(\text{ gof }{)}^{-1}=f^{-1}\text{o}{g}^{-1}$

Given $f:A\to B$ and $g:B\to C$ are bijective functions then we have $\text{ gof }:A\to C$ is also a bijection

$\Rightarrow (gof{)}^{-1}:C\to A$ a function and $g^{-1}:C\to B\text{ and }{f}^{-1}:B\to A$

$\Rightarrow f^{-1}o{g}^{-1}:C\to A$

$\therefore (gof{)}^{-1}$ and $f^{-1}o{g}^{-1}$ have same domain ‘C’ Let $c\in C$. since $g:B\to C$ is onto then there exist $b\in B$ such that $g\left(b\right)=c$

$\Rightarrow g^{-1}\left(c\right)=b$

Since $f:A\to B$ is onto for $b\in B$ then there exists

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

$\therefore \left(gof\right)\left(a\right)=g\left(f\right(a\left)\right)=g\left(b\right)=c$

$\therefore a=\left(gof{)}^{-1}\right(c)\dots \dots \left(1\right)$

$\left(f^{-1}o{g}^{-1}\right)\left(c\right)=f^{-1}\left(g^{-1}\left(c\right)\right)$

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

$\therefore \left(f^{-1}o{g}^{-1}\right)\left(c\right)=a\dots ..\left(2\right)$

Hence from $\text{ (1) \& (2) }(gof{)}^{-1}=f^{-1}o{g}^{-1}$