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 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 $0$,$1$ incomparable), thit 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 folows:

$f$ 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 that projections, (forgeting $k$ components), are onto maps as discribed.

I am wondering what kind of such onto mapping but projections does exist for some $d$ and $k\neq 0$.

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 image 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?