# There's an equivalence of simplicial categories $\Delta \to \tilde{\Delta}^{\text{op}}$.

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.

• Does $[0,1]$ mean $\{0,1\}$? Commented Apr 12, 2019 at 4:40
• 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? Commented Apr 12, 2019 at 5:16
• $[0, 1]$ means the totally ordered set $\{0,1\}$ where $0 \lt 1$. Commented Apr 12, 2019 at 7:15

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.

• How does this tie in with totally ordered posets? Commented Apr 17, 2019 at 21:17

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 $$\alpha_{\sigma}: \psi \circ \varphi(\sigma) \simeq \text{id}_{\Delta}(\sigma)$$.