$f:(X,\mathcal{T}_X)\mapsto (Y,\mathcal{T}_Y)$ is closed function if and only if for every topological space $(Z,\mathcal{T}_Z)$ and every continuous $1-1$ function $g:Y \mapsto Z$ the function $g \circ f:X \mapsto Z$ is closed function.
I have shown the inverse direction, but how one shows that $g(f(F))$ is a closed subset of $Z$?
It's not true. For instance take $X=Y=[0,2\pi)$, $f:x\mapsto x$ and $Z = S^1=\{z\in \mathbf C : |z| = 1\}$ and $g : x\mapsto e^{ix}$