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
> $\widehat{\square}$.
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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 ?
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**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](https://doi.org/10.1016/j.tcs.2009.11.013) and [Directed degeneracy maps for precubical sets](https://doi.org/10.48550/arXiv.2209.02667).


**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.
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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 $\widehat{\square}$.



Removing the symmetry maps from a small category of cubes
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 $\square$.
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**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 $f\in \square$.


