Let $f:A\to B,g:B\to C$ be bijection . Then show that $gof:A\to C$ is a bijection

Given $f:A\to B,g:B\to C$ be bijection.

To prove $gof:A\to C$ is a bijection then

We have to prove $gof:A\to C$ is One –one & onto

To prove gof is one-one :

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)

$g\left(f\left(a_1\right)\right),g\left(f\left(a_2\right)\right)\in C$

($\because g:B\to C$ is a function)

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

$\Rightarrow g\left(f\left(a_1\right)\right)=g\left(f\left(a_2\right)\right)$

$\because g:B\to C$ is one - one function

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

$\because f:A\to B$ is one – one function

$\therefore \text{ gof }:A\to C$ is one – one function

To prove gof is onto : 

Given $f:A\to B$ and $g:B\to C$ are two onto functions.

Let $c\in C$ since $g:B\to C$ is onto,there

Exists $b\in B$ such that $g\left(b\right) = c$

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

$a\in A$ such that $f\left(a\right)=b$

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

$\therefore$Every element of C has atleast one pre image in A

$\therefore \text{ gof }:A\to C$ is onto

$\therefore \text{ gof }:A\to C$ is one-one and onto

Hence $\text{ gof }:A\to C$ is bijection