Question

show that $f:A\to B$ and $g:B\to C$ are onto, then $gof:A\to$ is also onto:

Answer 1

Since $g:B\to C$ is on-to.

Let $aϵC$ then there will a pre-image of $a$ in $B$.

Let $P$ is the preimage.

$\therefore$ we can say $g\left(P\right)=a$.

Similarly,$f:A\to B$ is onto.

$\because PϵB$,then there is a pre-image in $A$.

Let $x$ be the pre-image of $P$

$\therefore f\left(x\right)=P\therefore gof=g\left(f\left(x\right)\right)gof=g\left(P\right)gof=a$

So,for every $x$ in $A$ there is an image in $C$.

So $gof$ is an onto function.