Let $\mathbf{Poset}$ denote the category of partially ordered sets and order-preserving maps. Does $\mathbf{Poset}$ have quotients?
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asked Oct 1, 2015 at 11:25
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(There may be some users who think this would have been better asked at Mathematics StackExchange, but I'll go ahead and answer because there are several ways of looking at it.)
The answer is "of course". See The Joy of Cats, p. 119. The coequalizer of two maps $f, g: X \rightrightarrows Y$ in $\mathbf{Poset}$ is computed in two steps: first take the coequalizer of $f, g$ as if in the category $\mathbf{Preord}$ of preordered sets (sets with reflexive transitive relations). This is done by taking the coequalizer in $\mathbf{Set}$, say $q: Y \to Q$, and then endowing $Q$ with the smallest reflexive transitive relation that makes $q$ an order-preserving map. Second, take the "posetal reflection of $Q$", i.e., the quotient $R$ where $x, y \in Q$ are identified if $x \leq y$ and $y \leq x$ in $Q$; the order relation on $R$ inherited from $Q$ makes $R$ the coequalizer of $f, g$ in $\mathbf{Pos}$.
In other language: $\mathbf{Preord}$ is topological over $\mathbf{Set}$ (see The Joy of Cats for details), and therefore is cocomplete (and much more). The full inclusion $\mathbf{Pos} \hookrightarrow \mathbf{Preord}$ is a reflective subcategory, and therefore $\mathbf{Pos}$ is also cocomplete, although not topological over $\mathbf{Set}$.
In still other language: it may be shown that $\mathbf{Preord}$ and $\mathbf{Pos}$ are locally finitely presentable, i.e., they are categories of structures defined entirely in terms of suitable finite limit diagrams in $\mathbf{Set}$, and on general abstract grounds such categories are cocomplete. See Adámek & Rosický, Locally presentable and accessible categories, Cambridge University Press 1994.
answered Oct 1, 2015 at 13:34
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1$\begingroup$ The need to go an extra step by taking a posetal reflection can be seen in simple examples. For example, if $Y$ is the poset which looks like $(a \to b\;\;\; c \to d)$ and $X$ is the discrete poset $(p \;\;\; q)$, and if we define $f$ by $f(p) = a, f(q) = b$ and $g$ by $g(p) = d, g(q) = c$, then the preorder quotient looks like a loop on two points, hence not a poset. The posetal reflection of that collapses to a point. $\endgroup$ Commented Oct 1, 2015 at 14:03
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1$\begingroup$ I do not understand how "there are several ways of looking at it" makes the problem research-level, or demonstrates that OP put effort into the question before asking it. I am downvoting this answer as it encourages more non-research-level questions on MathOverflow. $\endgroup$ Commented Oct 1, 2015 at 17:58
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1$\begingroup$ @BorisBukh I wasn't implying that that would make it "research level" (whatever that means exactly: it's such a relative term, cf. the fact that many logic questions get asked that a logician would consider elementary). The appropriateness for MO I leave for others to judge for themselves; I think it's plausible that this is something that a graduate student or non-expert mathematician might ask of another, in keeping with the original vision for MO. Also, while you might be right that not much effort went into the question, I did foresee that something interesting could be said. $\endgroup$ Commented Oct 1, 2015 at 18:38
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3$\begingroup$ @BorisBukh I would likely not have answered if there were already votes to close at the time of writing. In this case, the "amount of effort" can only be a guess. Maybe you're "right" in your judgment call; I don't know. But I respect the decision you made. If you want to continue the discussion, I suggest meta or chat, or you can mail me or the moderators if you like. $\endgroup$ Commented Oct 1, 2015 at 19:04
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