Inverse of Increasing Bijection need not be Increasing

Theorem
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\phi: S \to T$ be a bijection which is increasing.

Then $\phi^{-1}: T \to S$ is not necessarily also increasing.

Proof
Proof by Counterexample:

Let $S := \mathcal P \left({\left\{{a, b}\right\}}\right)$.

Let $T := \left\{{1, 2, 3, 4}\right\}$.

From Subset Relation on Power Set is Partial Ordering, we have that $\left({S, \subseteq}\right)$ is an ordered set.

Clearly so is $\left({T, \le}\right)$ (although its ordering is in fact total, it is still technically an ordered set).

Let $\phi: S \to T$ be defined as:
 * $\phi \left({\varnothing}\right) = 1$
 * $\phi \left({\left\{{a}\right\}}\right) = 2$
 * $\phi \left({\left\{{b}\right\}}\right) = 3$
 * $\phi \left({\left\{{a, b}\right\}}\right) = 4$

By inspection it can be seen that $\phi$ is a bijection.

Also by inspection it can be seen that $\phi$ is increasing.

But note that while $2 \le 3$, it is not the case that $\left\{{a}\right\} \subseteq \left\{{b}\right\}$.

That is, $\phi^{-1}$ is not increasing.