isomorphism in ordered monoids

I read that a morphism $$\gamma : S \rightarrow T$$ is an isomorphism if there exists a morphism $$\Psi : T \rightarrow S$$ such that $$\gamma \circ \Psi = I(T)$$ and $$\Psi \circ \gamma = I(S)$$, where $$I$$ denotes the identity.

In the category of semigroups [monoids, groups, semirings], a morphism is an isomorphism if and only if it is bijective.

But after some lines, it was written that the second sentence above does not hold for morphisms of ordered monoids. In particular, if a set $$(M,\leq)$$ is an ordered monoid with order relationship $$\leq$$, the identity induces a bijective morphism from $$(M,=)$$ onto $$(M,\leq)$$ which is not in general an isomorphism.

I got lost with the above paragraph. What could it mean?

• It means that the inverse map (of sets) is not a morphism of ordered monoids. If you want another example consider topological spaces: a bijective continuous map is not necessarily a homeomorphism, because the inverse may be non-continuous. – Michal Adamaszek May 24 at 7:15

If you consider two ordered monoids $$M,N$$, then it is perfectly possible for there to be a bijective order-preserving homomorphism from $$M$$ to $$N$$ whose inverse does not preserve the order. That is all you're getting an example of here. This applies also to ordered sets. The two-point set with the interesting order $$0<1$$ admits order-preserving bijections from the discrete order in which $$0$$ and $$1$$ are not comparable and to the indiscrete order in which $$0<1$$ and $$1<0$$. But neither bijection admits an order-preserving inverse.