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Let $\Delta$ be the simplicial category, that is, the category of finite totally ordered sets and order-preserving maps. Let $\tilde{\Delta}$ be the subcategory where objects are those of $\Delta$ and morphisms are order-preserving maps that also preserve smallest & largest elements.

Let $\varphi : \Delta \to \tilde{\Delta}^{\text{op}}$ be the functor sending $\sigma \mapsto \text{Hom}_{\Delta}(\sigma, [0,1])$ with the following order induced: $f \leq g$ in $ \text{Hom}_{\Delta}(\sigma, [0,1])$ if and only if $f(i) \leq g(i)$ for all $i \in \sigma$. I have checked that this is in fact a functor and that $\varphi(f) : \varphi(\sigma) \to \varphi(\tau)$ is indeed a map that preserves smallest and largest elements.

I'm want to prove that $\psi : \tau \to \text{Hom}_{\tilde{\Delta}}(\tau, [0,1])$ is quasi-inverse to $\varphi$.

That is to say there is a natural isomorphism $\psi \circ \varphi \simeq \text{id}_{\Delta}$.


At the very least we need that $\psi\circ (\varphi(\sigma)) \simeq \sigma$ or in other words $|\psi\circ\varphi(\sigma)| = |\sigma|$ in set cardinality.

Let $\sigma = []$ be the empty totally ordered set which happens to be the initial object of $\Delta$.

Then $\varphi(\sigma) =\{*\}$. So, $\psi(\{*\}) = []$ since there is no map from $\{*\}$ to $[0,1]$ that preserves both smallest and largest.

Now assume it's true for all $|\sigma|$ up to $n \in \Bbb{N} \cup \{0\}$. Then how would I show via induction that it's true for all $|\sigma| = n+1$?


If $|\sigma| = n + 1$, then a map $h: \sigma \to [0,1]$ restricted to $n$ elements of $\sigma$ is either $h(\sigma\setminus \{*\}) = 0$ in which case $h(\{*\})$ must be $1$, otherwise $h(\sigma \setminus \{*\}) = [0, 1]$ in which case $h$ so restricted is $\varphi(\sigma\setminus \{*\})$ and we know by induction that $\psi\circ\varphi(\sigma \setminus \{*\}) \simeq \sigma \setminus \{*\}$, and also that $h(\{*\}) = 1$.

Thus $\varphi(\sigma) = \{ h_0\} \cup \{h : h$ restricted to $\sigma\setminus \{*\} $ is in $\varphi(\sigma\setminus \{*\})$ and $h(*) = 1\}$.

Got that far so far.

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  • $\begingroup$ Does $[0,1]$ mean $\{0,1\}$? $\endgroup$ Commented Apr 12, 2019 at 4:40
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    $\begingroup$ This seems awfully laboured. Why not show that for a finite totally ordered set of order $n$, say $\{1,2,\ldots,n\}$ then there are $n+1$ order-preserving maps to $\{0,1\}$ and that $n-1$ of them preserve the largest/smallest element? $\endgroup$ Commented Apr 12, 2019 at 5:16
  • $\begingroup$ $[0, 1]$ means the totally ordered set $\{0,1\}$ where $0 \lt 1$. $\endgroup$ Commented Apr 12, 2019 at 7:15

2 Answers 2

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This is just a consequence of Birkhoff duality between finite posets and finite $0,1$-distributive lattices.

There is an equivalence between $\mathbf{FinPos}$ and $\mathbf{FinDistLat}^{op}$: to construct a distributive lattice from a finite poset, take all lower sets (including the empty one). To construct a poset from a finite distributive lattice, take all join-irreducibles and order them as in the lattice. Both constructions can be beefed up to be (contravariant!) functors.

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  • $\begingroup$ How does this tie in with totally ordered posets? $\endgroup$ Commented Apr 17, 2019 at 21:17
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Let $[0, ..., n]$ be a totally ordered set. Then the number or ways of mapping into $[0,1]$ that are order and min-max preserving can be determined by the number of splits: $([0], [1,...,n]), ([0,1], [1,...,n]), ..., ([0,1,..., n-1], [n])$ which is $n$.

You do the same "splitting" trick with $([], [0, ..., n]), ..., ([0,...,n], [])$ for just the order-preserving maps to get $n + 1$.


Thus if $\sigma = [1, ...., n]$ then $|\varphi(\sigma)|= n+1$ so that $\varphi(\sigma)$ is a totally ordered set $[0^*, 1^*, ..., n^*]$

By the first paragraph, $\psi$ maps us back to a totally ordered set of size $n$.

Thus there is always a unique isomorphism of totally ordered-sets $\alpha_{\sigma}: \psi \circ \varphi(\sigma) \simeq \text{id}_{\Delta}(\sigma)$.

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