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 $\widehat{\square}([1]^2,[1]^2)$:

- the identity $f(x_1,x_2)=(x_1,x_2)$
- the permutation $f(x_1,x_2)=(x_2,x_1)$
- $\gamma_1(x_1,x_2)=(\max(x_1,x_2),\min(x_1,x_2))$
- $\gamma_2(x_1,x_2)=(\min(x_1,x_2),\max(x_1,x_2))$.

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 $\widehat{\square}$.

sets, so it’s very confusing to see it meaning a small cube category? $\endgroup$1more comment