How to describe this set of maps of posets? 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 $f(\epsilon_1,\dots,\epsilon_n)=(\epsilon_{\sigma(1)},\dots,\epsilon_{\sigma(n)})$.

Is there such a representation theorem when $f$ is not bijective ?

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.
 A: I have found a way published in a recent preprint (https://doi.org/10.48550/arXiv.2209.02667).

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 $1\leq i\leq n$.

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):

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 $f_3(x_1,x_2,x_3)=\max(\min(x_1,x_2),\min(x_1,x_3),\min(x_2,x_3),\min(x_1,x_2,x_3))$.
