Bijection on Preordered Sets Implies Homeomorphism Prove that if $X$ and $Y$ are finite, then the "converse" of one of my other questions Homeomorphism on a Preordered Set
is true: 

if $h: X \to Y$ is bijective and satisfies  $\forall a,b \in X, \quad  (a \trianglelefteq_{\mathscr{S}}b \iff h(a) \trianglelefteq_{\mathscr{T}}h(b))$,
then $h$ is a homeomorphism.

We are assuming that $h$ is bijective, i.e., both one-to-one and onto. Our goal is to arrive at the conclusion that $h$ is continuous and has a continuous inverse in order to label it as a homeomorphism. 
I read somewhere that because $X$ and $Y$ are finite, then one-to-one and onto are equivalent, but I do not want to claim this and call it a day. Instead, I would like a more detail-oriented approach:
Since $h$ satisfies $a \trianglelefteq_{\mathscr{S}} b \iff h(a) \trianglelefteq_{\mathscr{T}} h(b)$, then if $a \in \overline{\{b\}}$ in $\mathscr{S}$, then $h(a) \in \overline{h(b)}$ in $\mathscr{T}$. So a closed set in one topology gets mapped to another closed set in a separate topology. Wouldn't this show that $h$ is continuous? 
As for the pre-image, could I apply $h^{-1}$ to the condition that is satisfied to get $h^{-1}(a) \trianglelefteq_{\mathscr{S}} h^{-1}(b) \iff a \trianglelefteq_{\mathscr{T}} b$ since $h$ is bijective, i.e., it has an inverse? 
Thank you in advance for reading this post and for any assistance provided, it is greatly appreciated. 
 A: A function $f:X\to Y$ with the property that $f[C]$ is closed in $Y$ whenever $C$ is closed in $x$ is said to be closed; it need not be continuous. In any case, you haven’t proved that $f$ has this property: you’ve merely shown that $h[\operatorname{cl}\{b\}]\subseteq\operatorname{cl}\{h(b)\}$ for each $b\in X$. Your result is an inclusion, not an equality, and it’s only for the closures of singletons, not for closed subsets of $X$ in general. (Note, by the way, that you should be writing $\overline{\{h(b)\}}$, not $\overline{h(b)}$: closure is an operation on sets, not on points.)
I would go back to the work that you did in connection with this question. For each $x\in X$ let $U_x=\{z\in X:x\trianglelefteq_{\mathscr{T}}z\}=\bigcap\{U\in\mathscr{T}:x\in U\}$, and for $y\in Y$ define $V_y$ analogously. You already know that $\mathscr{U}=\{U_x:x\in X\}$ is a base for $\mathscr{T}$ and that $\mathscr{V}=\{V_y:y\in Y\}$ is a base for $\mathscr{S}$. To show that $h$ is continuous, just show that $h^{-1}[V_y]\in\mathscr{T}$ for each $y\in Y$; this is pretty straightforward once you determine exactly what the set $h^{-1}[V_y]$ is.
To show that $h^{-1}$ is continuous, you want to show that $(h^{-1})^{-1}[U_x]\in\mathscr{S}$ for each $x\in X$, i.e., that $h[U_x]\in\mathscr{S}$ for each $x\in X$. This is really just like the previous argument.
