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In $Pos$ (the category of partial ordered sets and order preserving maps) there is the categorical product of two objects, but on the set product there is (naturally) also the lexicographic order. I ask:

has this latter some kind of categorical universal property? Or a categorical (external) characterization?

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  • $\begingroup$ I believe it's normally spelled "lexicographic" $\endgroup$
    – David White
    Commented Dec 8, 2011 at 16:30
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    $\begingroup$ Notice that the lexicographic order depends on which poset you choose to list first. Thus if it is to satisfy a universal property, this property must also depend on the order of the two posets. $\endgroup$
    – Manny Reyes
    Commented Dec 8, 2011 at 17:09
  • $\begingroup$ Manny Reyes: sure, but isnt essential, anyway consider the "first" (the primary for the lexicographic order) as the one on the left. $\endgroup$
    – Buschi Sergio
    Commented Dec 8, 2011 at 19:04
  • $\begingroup$ I seems resolved it, need preorders: the lexicographic preorder $x\times_l y$ is the universal object for the preorders $a$ by two morphism $f: a\to x,\ g: a\to |y|$ (where $|y|$ is the underling set of $y$ (caotic preorder)) plus a third morphism $h: a'\to y$ (where $a'$=*"pullback of $f$ by $\Delta_x\to x$"*) such that $\eta\circ h=g\circ \alpha$ where $\alpha: a'\to a,\ \eta: y\to |y|$ naturals $\endgroup$
    – Buschi Sergio
    Commented Dec 8, 2011 at 20:12
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    $\begingroup$ @Buschi - you are allowed to add an answer to your own question. You get more space and can format things better that way ;) $\endgroup$
    – David Roberts ♦
    Commented Dec 8, 2011 at 22:51

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As David Roberts said, I post the solution for preorders:

for two preorders $(X,\leq),\ (Y,\leq)$, let $(X,\leq)\times_l(Y,\leq)$ the lexicographic preorder (priority to the first factor $(X,\leq)$).

We observe that a preorder morphism $(A,\leq)\to (X,\leq)\times_l (Y,\leq)$ is a set map $(f, g): A\to X\times Y$ such that $f: (A,\leq)\to (X,\leq)$ is a preorder morphism, and $g: A\to Y$ a set map (we can view it as a morphism $g: (A,\leq)\to ch(Y)$ where $ch(Y)$ is the chaotic preorder on the set $Y$ (dont exists the functorial analogy for orders)) and If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$.

Reciprocally if we have the data maked as: a couple of morphisms like $f: (A,\leq)\to (X,\leq)$, $g: (A,\leq)\to ch(Y)$ such that "If $a\leq a'$ and $f(a)=f(a')$ then we have $g(a)\leq g(a')$", then we get the (unique) morphisms $(f, g): (A,\leq)\to (X,\leq)\times_l (Y,\leq)$.

Let $U: Preord\to Set$ the forgetfull functor from the preorders to sets category, this has a right adjoint funtor $Ch: Set\to Preord$ (the chaotic preorder), now the phrase "$a\leq a'$ and $f(a)=f(a')$" is described by the categorical relation $r_l: R_A(\leq)\cap P\subset A\times A$ where $R_A(\leq)$ is the preorder relation on $A$ and $P$ the pullback of $f\times f$ by the diagonal morphism $\Delta_X\subset X\times X$. Then $X\times_lY$ is universal (initial) for the objects $A$ with two morphisms $f: A\to X$, $g: A\to Ch(U(Y,\leq))$ such that $(g\times g)\circ r_l: R_A(\leq)\cap P\to Y\times Y $ as a (unique) factorization to the preorder relation $R_Y(\leq)\subset Y\times Y$ of $(Y,\leq)$.

now I realized it was just an easy exercise in translation of categorical logic, that is from the the formula to categorical diagrams.

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