Why are the isomorphisms and bijective morphisms not identical in the category of Pos?

Let $\text{Pos}$ be the category of partially ordered sets and monotonic functions. A morphism $f$ is called an isomorphism if there is a morphism $g$ such that $f\circ g$ and $g\circ f$ are identity morphisms. On the other hand a bijective morphism satisfies this condition,while it has been asserted that they are not identical. Can you explain what is happening? thanks in advance.

Take two elements, $0$ and $1$, and define the poset $X$ with $0\leq 1$ (and the reflexivity conditions), and $Y$ the poset only with $0\leq 0$ and $1\leq 1$ (i.e., we can't compare $0$ and $1$ in $Y$. Then the identity $Y\to X$ is a bijective morphism, but not an isomorphism (what would be it's inverse, and why is it not a morphism?).
If we were in the category of totally ordered sets, then any bijective morphism is in fact an isomorphism. What We did above was just take a simple poset ($X$) and weaken its structure a little bit (and obtain $Y$).
• @user318848 Of course every monotonic functions is a morphisms of $\mathrm{Pos}$ : this is the definition of a morphism in the category of posets. Your error lied in the belief that the set-theoretic inverse of a monotonic function is necessarily a monotonic function. – Pece Mar 1 '16 at 10:17