Let $n\geq 1$. Let $[n]=\{0<1\}^n$ equipped with the product order. Let $f:[n]\to [n]$ be a strictly increasing map. When $f$ is bijective, there exists a permutation $\sigma$ of $\{1,\dots,n\}$ such that $f(\epsilon\_1,\dots,\epsilon\_n)=(\epsilon\_{\sigma(1)},\dots,\epsilon\_{\sigma(n)})$.



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> Is there such a representation theorem when $f$ is not bijective ?
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For $n=2$, the only maps are $(\epsilon\_1,\epsilon\_2)\to (\epsilon\_1,\epsilon\_2)$, $(\epsilon\_1,\epsilon\_2)\to (\epsilon\_2,\epsilon\_1)$, $(\epsilon\_1,\epsilon\_2)\to (\min(\epsilon\_1,\epsilon\_2),\max(\epsilon\_1,\epsilon\_2))$ and $(\epsilon\_1,\epsilon\_2)\to (\max(\epsilon\_1,\epsilon\_2),\min(\epsilon\_1,\epsilon\_2))$. For $n\geq 3$, things become more complicated and I am not aware of any canonical representation, or at least of a way of listing all strictly increasing maps $f:[n]\to [n]$.


**Motivation**: When $f$ is bijective or for $n=2$, I can see $f$ as a continuous map $[0,1]^n\to [0,1]^n$ which moreover, for people interested in directed homotopy theory, takes a directed path of $[0,1]^n$ to another directed path of $[0,1]^n$ preserving the initial and final states of the $n$-cube. I would like to do the same thing for the other cases.



How to describe this set of maps of posets?
I have found a way published in a recent preprint (<https://doi.org/10.48550/arXiv.2209.02667>).



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> **Theorem**: Let $n\geq 1$. Let $f=(f\_1,\dots,f\_n):[n]\to [n]$ be a stricly increasing map. Then there is the equality $
> 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 $1\leq i\leq n$.
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I explain the result with an example instead. Consider the map $f:[3]\to[3]$ depicted in the following picture (the top diagram is the source, the bottom diagram is the image):


![](https://i.stack.imgur.com/w5pwy.png)
Let $f=(f\_1,f\_2,f\_3)$. For boolean values, $\min$ means "and" and $\max$ means "or". If $x\_1=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_1(x\_1,x\_2,x\_3)=1$. Thus $f\_1(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_3),\min(x\_1,x\_2,x\_3))$. If $x\_1=1$ and $x\_2=1$, or $x\_2=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_2(x\_1,x\_2,x\_3)=1$. Thus $f\_2(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_2),\min(x\_2,x\_3),\min(x\_1,x\_2,x\_3))$. Finally, if $x\_1=1$ and $x\_2=1$, or $x\_1=1$ and $x\_3=1$, or $x\_2=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_3(x\_1,x\_2,x\_3)=1$. Thus $f\_3(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_2),\min(x\_1,x\_3),\min(x\_2,x\_3),\min(x\_1,x\_2,x\_3))$.


