Question

If the mapping $f:A\to B$ and $g:B\to C$ are both bijective, then show that the mapping $gof:A\to C$ is also bijective.

Answer 1

Given $f:A\to B$ is bijective
$\Rightarrow f$ is one-one and onto.
Also, given $g:B\to C$ is bijective
$\Rightarrow g$ is one-one and onto.
Now, we will check whether $gof$ is bijective
Let $x,y\in A$ such that
$\left(gof\right)\left(x\right)=\left(gof\right)\left(y\right)$
$\Rightarrow g\left(f\right(x\left)\right)=g\left(f\right(y\left)\right)$
$\Rightarrow f\left(x\right)=f\left(y\right)$ (Since, g is one-one, so $g\left(x\right)=g\left(y\right)\Rightarrow x=y$
$\Rightarrow x=y$ ($\because f$ is one-one)
Hence, $gof$ is one-one.
Now, for surjective, let $z\in C$ be an arbitrary element
Since, g is onto, so for $z\in C$, there exists an element $r\in B$ such that $g\left(r\right)=z$
Also since, f is one so for every $x\in A$, there is an element $r\in B$ such that $f\left(x\right)=r$
$\Rightarrow g\left(f\right(x\left)\right)=g\left(r\right)=z$
$\Rightarrow \left(gof\right)\left(x\right)=z$
Hence, for $z\in C$, there is an element $x\in A$ .
Hence, $gof$ is onto.