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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.

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

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