Let $S = \{1,2,3,4\}$. Let $F$ be the sets of all functions from $S$ to $S$.
Now I think this statement is true:
$\forall f \in F , \exists g\in F$ so that $(g \circ f)(1) =2$
I suppose $f \in F$, and I need to show that $\exists g\in F$ so that $(g \circ f)(1) = 2$
Do I just make an example of $g$ so that $(g \circ f)(1) = 2$ ?
Like $f(1) = y$ and then $g (y)=2 ?
You could simply design a function $g$ such that $g(x)=2$ for all $x\in S$. Now since $f\in F$, we know $f(1)\in S$, hence the composition is 2. As f is arbitrary, we have the desired result.