Let $S = \{1,2,3,4\}$. Let $f$ be the sets of all functions from $S$ to $S$.
Now, is this statement correct?
$\forall f \in F$ , $\exists$ $g \in F$ so that $(f\circ g) (1) = 2$
I think this is false because suppose the only output $f(x)$ can produce is
$f(x)=3$ and $f(x) = 4$
Now, there is no possibility of $(f\circ g)(1) = 2$ because $2$ isn't an output of $f$.
Is that correct to say?
Yes, this is correct. Perhaps easier is to consider $f(x)=1$ for all $x$, but the idea is the same.