Is it reasonable to define `poset homotopy' as a `natural transformation of posets'?

Question

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?)

Answer 1

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.

Answer 2

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.