We are given two functions f : A → B and g : B → C. Prove that if f and g are onto, then g ◦ f is onto.

Since $g: B \rightarrow C$ is onto

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

Let the pre-image be y

Hence, $y \in B$ such that $g(y)=z$

Similarly, since $f: A \rightarrow B$ is onto

If $y \in B$ , then there exists a pre-image in A

Let the pre-image be $x$

Hence, $x \in A$ such that $f(x) =y$

Now,

$\begin{aligned} \text { gof : } A \rightarrow C \\ \begin{aligned} \text { gof } &=g(f(x)) \\ &=g(y) \\ &=z \end{aligned} \end{aligned}$

So, for every $x$ in A, there is an image $z$ in C . Thus, $gof$ is onto.