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Does the category of partial orders have a subobject classifier? (Edit: No, see Eric's answer.)

If not, what is a category which is "close" to the category of partial orders (e.g. it should consists of special order-theoretic constructs) and has a subobject classifier? Bonus question: Is there also such an elementary topos? Notice that the category of partial orders has all limits, colimits and it is cartesian closed.

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    $\begingroup$ A cartesian closed category with equalisers and subobject classifiers is an elementary topos, but $\mathbf{Poset}$ is not. $\endgroup$ – Zhen Lin Feb 11 '16 at 8:09
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The category of posets has no subobject classifier. Indeed, if there did exist a universal subobject $1\to \Omega$, then it would be a regular subobject, and so every subobject would be regular since regular subobjects are stable under pullback. But the regular subobjects are exactly the "subposets" in the usual sense (i.e., subsets with the inherited order), and so not all monomorphisms of posets are regular (because the domain can have a weaker order than the codomain).

[In fact, if you work with "evil" posets (i.e., antisymmetric preorders), there are even regular subobjects which cannot be pulled back from any terminal subobject (here by "terminal subobject" I mean a subobject $A\to B$ where $A$ is a terminal object, as a universal subobject $1\to\Omega$ would have to be). For instance, it is easy to see that if $P$ is a poset and $Q\subseteq P$ is the pullback of a terminal subobject $1\to R$ via some map $P\to R$, then for all $a,b\in Q$ and all $c\in P$, $a\leq c\leq b$ implies $c\in Q$.]

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  • $\begingroup$ Thank you! I suspected this. Do you have an idea for my extended question? $\endgroup$ – Martin Brandenburg Feb 11 '16 at 8:43
  • $\begingroup$ For instance, your argument doesn't seem to work for preordered sets, so one might hope for a subobject classifier for them. $\endgroup$ – Martin Brandenburg Feb 11 '16 at 9:04
  • $\begingroup$ I don't have any immediate thoughts on your extended question. I have updated the answer to give an argument that works for preorders. $\endgroup$ – Eric Wofsey Feb 11 '16 at 9:13
  • $\begingroup$ Thank you. But why is $1 \to \Omega$ regular? $\endgroup$ – Martin Brandenburg Feb 11 '16 at 9:22
  • $\begingroup$ There might be some general abstract argument, but the argument I had in mind is simply that a singleton set can only be given one ordering. $\endgroup$ – Eric Wofsey Feb 11 '16 at 9:26
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The fact that all internal co-categories in a coherent category are necessarily co-equivalence relations [see Peter Lumsdaine's TAC article A small observation on co-categories] provides a telltale sign that the category of posets fails to be a topos.

For the inclusion functor $\textbf{Poset} \to \textbf{Cat}$ is represented by the internal co-category whose (co-?)nerve is the inclusion of the non-empty finite ordinals $\Delta \to \textbf{Poset}$, which is evidently not a co-equivalence relation (since not all posets are equivalence relations).

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