I'm studying category theory for the first time in a very succint book for computer scientists (I'm not actually a computer scientist, I'm a physicist, but my interest in cat theory is related to purely functional programming languages). But, as the book is very succint, may be it lacks some information so I have a question.

Supose I have a category with a finite number of objects {a1, a2, ..., an}. Supose also that, if there is a morphism f : ai -> aj connecting two objects, then it is unique. Does this category always correspond to some partial order in the set {a1, a2, ..., an}?

I convinced myself drawing some diagrams that this could be the case (couldn't find a counter example), but I'm not sure.

edit: I see your point. I don't have antisymmetry garanteed.

Suppose I further restrict things: I'm not going to allow a morphism (b -> a) it there's already a (a -> b). Then I fix this, right? It'll be a partial order with no equalities, right?


2 Answers 2


Almost, this has nothing to do with finiteness: any category where the homsets have at most one element each is a preorder. Define $a\le b$ if there is an arrow from $a$ to $b$. Then $\le$ is reflexive and transitive, by the category axioms. But it may fail to be antisymmetric: one may have $a\le b\le a\ne b$. But a preorder induces a partial order on the equivalence classes with respect to the relation $a\sim b$ if $a\le b\le a$: in the category these are isomorphism classes of elements.


Not quite, you may have two distinct objects in the category that are isomorphic so what you get is a relation which does not have $x\leq y\wedge y\leq x\implies x=y$. However you get a well-defined partial order on the set of isomorphism classes of objects.

  • $\begingroup$ Hummmm I see... So if I restrict even more, say, not allowing a morphism (b-> a) if there is already a morphism (a -> b), I do always get a poset with strict inequalities (no equalities), right? $\endgroup$
    – Rorsa
    Jul 27, 2010 at 20:52
  • 1
    $\begingroup$ Well, you can't do that if $a=b$. But if $a\ne b$ if you disallow both $a\to b$ and $b\to a$ holding simultaneously, you get a partial order. $\endgroup$ Jul 27, 2010 at 20:57

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