Question

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

Answer 1

Since $h:B\to C$ is onto

Suppose $z\in C$, then there exists a rpe-image in $B$

Let the pre-image be $y$

Hence, $y\in B$ such that $g\left(y\right)=z$

Similarly since $f:A\to B$ is onto

If $y\in B$ then exists a preimage in $A$

Let the pre image be $x$

Hence, $x\in A$ such that $f\left(x\right)=y$

Now, $g\circ f=g\left(f\right(x\left)\right)$

$=g\left(y\right)=z$

So, for every $x$ in $A$, there is an image $z$ in $C$, thus $g\circ f$ is onto.