Let $[1]^n=\{0<1\}^n$ equipped with the product order. I consider the small category $\widehat{\square}$ of the category of partially ordered sets generated by the coface maps $\delta^\epsilon_i:[1]^{n-1}\to [1]^n$ with $\epsilon=0,1$ defined by $$\delta^\epsilon_i:(x_1,\dots,x_{n-1}) \mapsto (x_1,\ldots,x_{i-1},\epsilon,x_i,\dots,x_{n-1})$$ and by the strictly increasing maps $f:[1]^n\to [1]^n$.

The small category $\widehat{\square}$ contains the symmetry maps (the ones permuting the coordinates).

Informally, I would like to remove the symmetry maps from $\widehat{\square}$, and only them, to obtain a subcategory of $\widehat{\square}$.

Every strictly increasing map $f=(f_1,\dots,f_n):[1]^n\to [1]^n$ gives rise to another strictly increasing map by permuting the coordinates. I need to find a way to make a choice among all permutations.

Every strictly increasing map $f=(f_1,\dots,f_n):[1]^n\to [1]^n$ satisfies the equalities $$f_i(x_1,\dots,x_n) = \max_{(\epsilon_1,\dots,\epsilon_n)\in f_i^{-1}(1)} \min \{x_k\mid \epsilon_k=1\}$$ for all $1\leq i\leq n$ (see https://mathoverflow.net/a/429941/24563).

Question: In the formula above, is there a way to put a total order on the coordinates by using the syntax of the formula ?

Motivation: I work with the presheaves on $\widehat{\square}$ that I call transverse sets. They are a generalization of the category of precubical sets adapted for studying the directed homotopy for concurrency. And I would like to define the non-symmetric transverse sets. The two papers using transverse sets are Combinatorics of labelling in higher dimensional automata and Directed degeneracy maps for precubical sets.

EDIT (I add some details to give some intuition): a (too) naive idea consists of defining this subcategory of $\widehat{\square}$ by using this lemma:

Fact: Every map $f:[1]^m\to [1]^n$ of $\widehat{\square}$ factors uniquely as a composite $[1]^m\to [1]^m \to [1]^n$ where the right-hand $[1]^m\to [1]^n$ is a composite of coface maps.

And then to consider the subset of maps of $\widehat{\square}$ factorizing like $[1]^m\to [1]^m \to [1]^n$ such that the left-hand map is not one-to-one unless it is the identity of $[1]^m$ and such that the right-hand map is a composite of coface maps. Unfortunately, this subset of maps of $\widehat{\square}$ is not closed under composition. Here is a simple counterexample.

  • $f:[1]^2\to [1]^4$ defined by $f(x_1,x_2)=(x_1,x_2,0,0)$
  • $g:[1]^4\to [1]^4$ defined by $g(x_1,x_2,x_3,x_4) = (x_2,x_1,\max(x_3,x_4),\min(x_3,x_4))$.

$f$ is a composite of coface maps. $g$ is not one-to-one since $$g(x,x,1,0)=g(x,x,0,1)=(x,x,1,0)$$ for $x=0$ or $x=1$. However $$(g\circ f)(x_1,x_2)=(x_2,x_1,0,0)$$ which means that $g\circ f$ is the composite of a nontrivial permutation map $[1]^2\to[1]^2$ followed by a composite of coface maps.

There are four maps in $\widehat{\square}([1]^2,[1]^2)$:

  1. the identity $f(x_1,x_2)=(x_1,x_2)$
  2. the permutation $f(x_1,x_2)=(x_2,x_1)$
  3. $\gamma_1(x_1,x_2)=(\max(x_1,x_2),\min(x_1,x_2))$
  4. $\gamma_2(x_1,x_2)=(\min(x_1,x_2),\max(x_1,x_2))$.

The idea would be to keep from $\widehat{\square}([1]^2,[1]^2)$ the identity and one of the two maps crushing the square transversally $\gamma_1$ or $\gamma_2$, and to find a way to do the same thing for all sets $\widehat{\square}([1]^m,[1]^n)$ in such a way that we obtain a subcategory of $\widehat{\square}$.

  • $\begingroup$ By "posets" you mean finite sets? And how do you get symmetries in that category? And isn't the only strictly increasing map $f : [n] \to [n]$ the identity? $\endgroup$ Commented May 19, 2023 at 1:59
  • $\begingroup$ @darijgrinberg $[n]=\{0<1\}^n$, not $\{1<\dots<n\}$. For example, the map $(x_1,x_2)\mapsto (x_2,x_1)$ is strictly increasing. Poset means partially ordered set. $\endgroup$ Commented May 19, 2023 at 6:20
  • $\begingroup$ This is a nice question, but may I suggest two notational changes might make it clearer? Denoting the cube posets $[1]^n$ or $I^n$ rather than $[n]$ (both certainly occur in the cubical sets literature, but elsewhere $[n]$ is much more widely used for $\{0,…,n\}$, which the cubical sense clashes badly with); and using something like $\square'$ instead of $\widehat{\square}$, since $\widehat{-}$ is standard notation for presheaf categories, and in particular $\widehat{\square}$ is commonly used for the category of cubical sets, so it’s very confusing to see it meaning a small cube category? $\endgroup$ Commented May 19, 2023 at 7:50
  • $\begingroup$ @PeterLeFanuLumsdaine Probably $[1]^n$ is better than $[n]$ indeed. I use a different notations for presheaves. $\endgroup$ Commented May 19, 2023 at 8:26
  • $\begingroup$ Ah, those are the $[n]$s here! Interesting, but yes, the notation should be better... $\endgroup$ Commented May 19, 2023 at 13:56

1 Answer 1


The naive idea has to be slightly modified. The point is not to sort out all terms (it is a wrong intuition), but only where the variables $x_i$ are "alone". For example, the map $$(x_1,x_2,x_3,x_4)\mapsto (x_2,x_1,\max(x_3,x_4),\min(x_3,x_4))$$ is not kept in the subcategory because $x_2$ which is alone is before $x_1$ which is alone. On the contrary, the map $$(x_1,x_2,x_3,x_4)\mapsto (x_1,x_2,\max(x_3,x_4),\min(x_3,x_4))$$ will be kept. This way, symmetry maps cannot show up by precomposing by coface maps, and the counterexample of the question cannot exist. Everything boils down to the following proposition ($\square$ is the box category, which is generated by the coface maps):

Proposition: The set of maps $$\mathcal{A}=\{\phi:[1]^m\to[1]^n\in \widehat{\square}\mid \forall \delta:[1]^p\to [1]^m\in \square, \phi\delta \hbox{ one-to-one }\Rightarrow \phi\delta\in \square\}$$ is closed under composition and contains all identity maps. Moreover, the only one-to-one maps of $\mathcal{A}$ are the maps of $\square$.

Proof: Let $\phi_1,\phi_2\in \mathcal{A}$ such that $\phi_1\phi_2$ exists. Let $\delta\in \square$ such that $\phi_1\phi_2\delta$ exists and is one-to-one. Then $\phi_2\delta$ is a one-to-one set map. Thus $\phi_2\delta\in \square$, $\phi_2$ belonging to $\mathcal{A}$. We deduce that $(\phi_1\phi_2)\delta =\phi_1(\phi_2\delta) \in \square$ since $\phi_1\in \mathcal{A}$. This means that $\phi_1\phi_2 \in \mathcal{A}$ by definition of $\mathcal{A}$. $\mathcal{A}$ contains clearly the identity maps. Finally suppose that $f:[1]^m\to [1]^n\in \mathcal{A}$ is one-to-one. Then $f\mathrm{id}_{[1]^m}$ is one-to-one, which implies that $f\in \square$.


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