Discrete math surjective function proof

Question

so the exercise goes:

Let $E, F$ and $G$ be given sets and let the functions $f: E \to F$ and $g: F \to G$ be given. Consider the claim:

if $f$ is surjective and $g$ is surjective $\implies g\circ f$ is surjective.

The claim is either true or false. If it is true, prove it. If it is false, give example of 3 sets of $E,F,G$ and two functions $f,g$ where the fuctions $f:E\to F$ and $g: F\to G$ are surjective but $g\circ f$ isn't surjective.

How should I think about all of this? We've had so few tasks in our course on this subject

Answer 1

A function $h: A \to B$ is surjective if and only if for all $y \in B$ there exists an $x \in A$ such that $h(x)=y$.

Thus, let us consider a generic $y \in G$. Since $g$ is surjective, by definition there exists $w \in F$ such that $g(w)=y$.

Also, since $f$ is surjective, there exists $x \in E$ such that $f(x)=w$.

So for a generic $y \in G$, we have guaranteed the existence of $x$ such that $g(f(x))=g(w)=y$.

Answer 2

A variant (more synthetic) proof:

$g\circ f$ is surjective means the image $(g\circ f)(E)=G$.

This is true because \begin{align} (g\circ f)(E)=g\bigl(f(E)\bigr)&=g(F)& &\text{($f$ is surjective)}\\ &=G&&\text{($g$ is surjective).} \end{align}