Denote by $Q\_n$ the $n$-dimensional hypercube. A vertex of $Q\_n$ is represented by a vector of $n$ $\{0,1\}$-bits. An edge corresponding to two vertices whose vectors differ in one coordinate is represented by a bit vector with a $\*$ in that coordinate. A $d$-dimensional subcube $D$ is represented by a bit vector with $d$ $\*$’s, where the vertices of $D$ are those vectors obtained by replacing the $\*$’s with any combination of bits. 


Now let's introduce the natural order relation on subcubes of hypercube $Q\_n$ by setting $x\leq y$ iff $x\_i\leq y\_i$ (where $\*\leq 0$ and $\*\leq 1$ and $0$, $1$ are incomparable). This poset of subcubes is identified with $\{\*,0,1\}^d$ and is a face-poset of a hypercube $Q\_d$.


Let me consider a map $f:S(Q\_{d+k})\to S(Q\_d)$ from face-poset of $Q\_{d+k}$ to face-poset of $Q\_d$ which is as follows:



> 
> $f$ preserves the order relation, i.e., if $x\leq y$ then $f(x)\leq f(y)$
> 
> 
> 


and 



> 
> if for $x\in Q\_{d+k}$ holds that $f(x)\leq z$ then there exist $y\in Q\_{d+k}$ such that $x\leq y$ and $f(y)=z$.
> 
> 
> 


M. Winter noticed below in [comments](https://mathoverflow.net/questions/338089/maping-of-subcubes-of-a-dk-hypercube-onto-subcubes-of-d-hypercube#comment845491_338089) that projections, (forgetting $k$ components), are onto maps as described.



> 
> I am wondering what other kind of onto mappings exist for some $d$ and $k\neq 0$. 
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> 
> 


A later addition:



> 
> In particular, I am wondering if it is possible to map a face-poset of $Q\_{3+k}$ to the face-poset of $Q\_3$ by onto mapping as described above in such a way that inverse images of $3$-face are all of the dimension greater then $3$, and how big the dimension of such inverse images can be.
> 
> 
> 


Is it true that the dimension of such inverse images is not bounded by any number by means of varying $k$, or the bound does exist?



Mapping of subcubes of a $(d+k)$-hypercube onto subcubes of a $d$-hypercube
Take any maximal face $a$ in the pre-image of $(\*,\*,\*)$ (i.e., of the 3-dimensional face). W.l.o.g., it has the form 
$a=(\underbrace{\*,\dots,\*}\_k,0,0,\dots,0)$. Denote by $a(p,q)$ the face obtained from $a$ by replacing the $p$th star with $q$ (so $1\leq p\leq k$ and $q\in\{0,1\}$).


We say that a bit is *determined* if it is not a star. Notice that each $f(a(p,q))$ has a determined bit.


**1.** Assume that $f(a(p,0))$ and $f(a(p,1))$ have some determined bits in different positions, say $f(a(p,0))$ has the first bit $0$, and $f(a(p,1))$ has the second bit $0$ (wlog $p=1$). Then there is no $y\geq a$ such that $f(y)=(1,1,\*)$. Indeed, if $y$ has a determined first bit, then $f(y)$ has $0$ in the corresponding bit. Otherwise, if $y$ has a star in the first bit, then replace it by $0$ to get $y\_0$. We have $f(y\_0)=(0,\dots)$, so we cannot have $f(y)=(1,\dots)$.


Thus, each of $f(a(p,0))$ and $f(a(p,1))$ has a unique determined bit, and those bits are at the same position $s(p)\in\{1,2,3\}$.


**2.** Assume that $f(a(p,0))$ and $f(a(p,1))$ coincide, say $f(a(1,0))=f(a(1,1))=(0,\*,\*)$. Then there is no $y\geq a$ with $f(y)=(1,\*,\*)$. The reasoning is similar: if the first bit of $y$ is determined, then that of $f(y)$ is $0$; otherwise, replace the first bit in $y$ by $0$.


Thus, $f(a(p,0))$ and $f(a(p,1))$ differ in exactly $s(p)$th bit, the others being stars.


**3.** Assume that the function $s$ is not injective, say $s(1)=s(2)=1$. We may assume that $f(a(1,0))=(0,\*,\*)$ and $f(a(2,1))=(1,\*,\*)$. Then there is no image for $(0,1,\underbrace{\*,\dots,\*}\_{k-2},0,\dots,0)$.


Thus $s$ is injective, which yields $k\leq 3$.


The same reasoning works when 3 is replaced with any other number.


