Originally Posted by TheMathsDude
Finding it hard to manipulate the surjective definition to fit the previous examples written.

Any clues?
i'll help you out, i leave the actual proof to you, but i'll give you somewhat of an outline.

Definition: A function $f$ is surjective (or onto) if for every $y$ in the codomain of $f$, there is an $x$ in the domain of $f$ such that $f(x) = y$

now what that basically means, is that the range of the function is the same as the codomain of the function.

so we have $f : A \to B$ and $g:B \to C$. so $g \circ f : A \to C$.

so, since $g \circ f$ is surjective, $C$ is the codomain and at the same time, the range of $g \circ f$. now what $g \circ f$ does, is use the function $f$ to take some element $x \in A$ and put it in $B$ and then use the function $g$ to take that element and then move it to $C$. this last part of the function is what makes it surjective, since it is the part that maps $x$ to the range (which is the codomain).