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Pausing to speak metaphorically for a bit, I have always thought of partial orders as being something like a simplified version of topological spaces. (In the finite case at least, all topological spaces are partial orders). This leads to thinking about continuous maps as a parallel concept to the idea of a poset homomorphism. But is there a corresponding parallel notion for a homotopy of two continuous maps?

Here is a sketch of how I imagine this working:

Since all preorders are categories, and posets are preorders, we can think of any partial order $(\leq, C)$ as a category whose morphisms are just given by the relation $\leq$. Now a homomorphism of posets is just a functor of these categories. Taking this one step further, we can define a natural transformation of any two homomorphisms of posets, which I am thinking of as a candidate definition for a poset homotopy.

Now it is also true that CW complexes are technically posets under the incidence relation. If we replace the posets in the above sketch with CW complexes and the functors with cellular maps, does the proposed definition for poset homotopy match the known definition for a `CW complex homotopy'? (EDIT: Or is this true for some variation of a CW complex; for example a finite or simplicial complex?)

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Re: your last paragraph: simplicial complexes are posets, but CW-complexes have more information than incidence, contained in their attaching maps. –  Mariano Suárez-Alvarez Apr 11 '11 at 19:26
@Mariano: Thanks, I updated the question to reflect your comment. –  Mikola Apr 11 '11 at 19:32
In the finite case, a complete partial order is just a poset with a least element. The usual definition of a poset from a finite topology is to declare x < y in the poset if the closure of {x} contains y in the topology. I don't see why this needs to have a least (or greatest) element. –  Dylan Thurston Apr 11 '11 at 19:51
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2 Answers

up vote 3 down vote accepted

This is an interesting line of questions, but I think it doesn't quite work as stated. First off, your notion of "homotopy" is not an equivalence relation (as far as I understand it), so it won't agree with a topological notion.

But there are also other issues; basically, any notion of "poset homotopy classes" along the lines you suggest will have finitely many classes of maps between two spaces, unlike real homotopy classes.

For instance, the poset $P = \{x,y,z,w\}$ with $x > z$, $x > w$, $y > z$, and $y > w$ corresponds to a finite topological space that is weakly homotopy equivalent to $S^1$, so $\pi_1(P) = \mathbb{Z}$. But there are only finitely many poset maps from $P$ to $P$, even before taking any homotopy equivalence.

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Thanks for the response! I am still in the process of trying to teach myself algebraic topology and end up spending a lot of time blundering around trying to relate it back to things that I already know. Maybe another way to do this construction would be to more directly parallel the usual definition by replacing the cartesian product with [0,1] by a cartesian product with a span; {0, t, 1} where 0 < t and 1 < t. Of course at this point I am probably grasping at straws and so I should perhaps just stop now and try to think a bit more before I say anything else... –  Mikola Apr 11 '11 at 20:29
Using posets is too restrictive. If one replaces posets by (small) categories, then "natural transformation" is an equivalence relation. Furthermore, with respect to this definition, the homotopy category of (small) categories is equivalent to the homotopy category of spaces. –  John Klein Apr 11 '11 at 23:38
@John: Natural transformations are not necessarily invertible, so the relation of "the existence of a natural transformation" is not symmetric. Also, it's true that if you invert the functors which induce weak homotopy equivalences of nerves, then you get a homotopy category equivalent to that of spaces, but I don't think that's the same thing as quotienting by the equivalence relation generated by natural transformations; not all objects in the Thomason model structure are fibrant and cofibrant. –  Mike Shulman Apr 14 '11 at 20:04
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For any finite poset $P$, we can consider the poset $sP$ of nonempty chains in $P$, ordered by inclusion. There is a morphism $m_P:sP\to P$ sending each nonempty chain to its largest element. If you modify the category of finite posets by

  1. Identifying maps $f,g:P\to Q$ whenever they are homotopic (ie $f(p)\leq g(p)$ for all $p\in P$); and

  2. Adjoining inverses for the maps $m_P$

then you get the usual homotopy category of finite complexes. I won't swear that the details are completely straight, but certainly something like this is true. Key points are that

  • We can regard $P$ as an abstract simplicial complex, where the simplices are the nonempty chains. We therefore have a geometric realisation $|P|$ (and these satisfy $|P\times Q|=|P|\times |Q|$).
  • An arbitrary abstract simplicial complex $K$ need not arise from a poset, but if we let $sK$ denote the poset of simplices in $K$ then $|sK|$ is homeomorphic to $|K|$ by barycentric subdivision.
  • If we write $I=\{0,1\}$ (ordered in the usual way) then $|I|$ is the unit interval. If $f,g:P\to Q$ with $f\leq g$ then we can define $h:I\times P\to Q$ by $h(0,p)=f(p)$ and $h(1,p)=g(p)$, and then $|h|$ gives a homotopy between $|p|$ and $|q|$.
  • The homeomorphism $|sP|\to |P|$ provided by barycentric subdivision is homotopic to $|m_P|$ (so in particular, $|m_P|$ is a homotopy equivalence).

The claim now follows from the simplicial approximation theorem. I think I first saw this kind of formulation in a book by Rourke and Sanderson.

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I believe the original reference is McCord, Duke Math. J. 33 (1966), 465-474. Here's another way to say it. There are two ways to get a topological space from a finite poset $P$. First we can simply let the downsets of $P$ (i.e., subsets $I\subseteq P$ such that $x\le y \in I$ implies $x\in I$) be the open sets of a topology on the ground set of $P$. The other way is to let the totally ordered subsets of $P$ define an abstract simplicial complex, and consider some geometric realization. McCord shows that there is a weak homotopy equivalence between these two topological spaces. –  Timothy Chow Apr 12 '11 at 21:38
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