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Is there a standard name for order-preserving maps $f\colon P\to Q$ of posets with the property that the image of a lower set is a lower set, or equivalently if $q\leq f(p)$ then there exists $p'\leq p$ with $f(p')=q$?

If you view the poset as a category, then this condition says that the functor associated to $P$ is surjective on in-stars.

In my research I need to consider the category of posets with these kinds of morphisms and I would like to know their name. For example, if $P$ and $Q$ are face posets of regular CW complexes, then this property says that the image of each closed cell is a closed cell.

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  • $\begingroup$ You don't also insist that the map is order-preserving? $\endgroup$
    – Joel David Hamkins
    Commented Jul 19, 2013 at 15:40
  • $\begingroup$ By maps of posets I meant order preserving. I'll edit. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 19, 2013 at 15:47
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    $\begingroup$ I suppose it could also be ambiguous whether one means $p\leq q\iff f(p)\leq f(q)$, or $p\leq q\to f(p)\leq f(q)$ or $p\lt q\to f(p)\lt f(q)$. $\endgroup$
    – Joel David Hamkins
    Commented Jul 19, 2013 at 16:19
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    $\begingroup$ @JoelDavidHamkins, by order preserving I mean $a\leq b$ implies $f(a)\leq f(b)$. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 19, 2013 at 16:37
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    $\begingroup$ @The Masked Avenger. Now that you mentioned it, it is actually a topological condition. If we give posets the topology where the lower sets are precisely the open sets, then the order preserving maps are precisely the continuous functions, and the open maps are precisely the maps that map lower sets to lower sets. Similarly, if we give each poset the topology where the lower sets are precisely the closed sets, then the closed maps are precisely the maps that map lower sets to lower sets. $\endgroup$
    – Joseph Van Name
    Commented Jul 19, 2013 at 21:22

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I think one name for this is a simulation.

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  • $\begingroup$ thanks for this. I will wait to see if poset people have a different name before accepting. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 20, 2013 at 5:47

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