# Are Morphisms of a Category Order Isomorphisms?

Let the objects be all partially ordered sets $$(S,\le)$$ in a Category $$\mathscr{C}$$. A morphism $$(S,\le) \to (T,\le)$$ is a function $$f: S \to T$$ such that for $$x,y \in S, x \le y \implies f(x) \le f(y)$$.

I have the statement above in my Reference Book (Algebra of Hungerford). It is quoted without proof and entitled as "Example", but if so then without enunciation. I don't see why any morphism class over categories could define an order isomorphism.

Is this a fact- and why? Otherwise thank you to show me the way to see it clearer.

• This is just a definition of a particular category, where morphisms are simply defined to be order-preserving maps. – asdq Jan 11 at 8:42
• @asdq Thank you – freehumorist Jan 11 at 8:52