Is any underlying Hypergraph of a finite cubical complex contained in the Hypergraph of cubical complex formed by n-cube,with n sufficient large?

Question

This question arises from simple observation that an oritentated finite simplicial complex must be contained in a sufficient large simplicial complex formed by n-simplex(Take n as number of vertexes in finite simplicial complex) But I don't know whether this property holds in cubical complex.

Essentially speaking what I want to prove is: Any cubical complex(orientated or not) has an underlying hypergraph structure, is it a sub-hypergraph of an underlying hypergraph of n-cube?

Or in topology aspects: there is a homeomorphism between any finite cubical complex in $\mathbb{R}^m$ and a sub cubical complex of n-cube complex in $\mathbb{R}^n$.

Supplementary details: For n-cube: it is an n-times Cartesian product of interval $[0,1]$ in $\mathbb{R}^m$

cubical complex: A cubical complex K in $\mathbb{R}^m$ is a collection of k-cubes in $\mathbb{R}^m$ such that

1. Every face of a cube of $K$ is in $K$, and

2. The intersection of any two cubes of $K$ is a face of each of them

(This definition varies with articles, here is an analogy of simplicial complex)

Underlying hypergraph: a finite cubical complex has finite vertexes V, with all of its cubes as hyperedge in the hypergraph.

Answer 1

Let the $n$ cube consist of the collection of $n$-tuples whose entries are each $0$ or $1$ And define the $m$ dimensional lattice with bound $k$ as the collection of $m$ tuples whose entries are each in the set $\{0,1,\cdots,k\}.$ Given any finite cubical complex it is, for large enough $m,k,$ a subcomplex of the $m$ dimensional lattice of bound $k.$ Note that the only vertices allowed to be connected in the bounded lattice are pairs of m tuples which differ in only one component, and that difference is by 1. For example $(1,5,2)$ may connect to $(1,6,2)$ but not directly to $(1,7,2).$

Our idea now is to embed the bounded $m$ dimensional lattice in an $n$ cube where $n$ is large enough. To do so we consider a particular point $(a_1,\cdots,a_m)$ of the bounded $m$ dimensional lattice. Encode it at first into a $0,1$ matrix for which each $a_k$ forms the $k$th row of the matrix as a sequence of $a_k$ 1's followed by all 0's. The matrix formed, map it into the $n$ cube by placing the rows of the matrix side by side, from first row to last.

The subcomplex of the n cube formed this way will have the same vertex pairs connected as in the m dimensional lattice with bound k, namely two vertices are connected iff they differ in only one component, and that component difference is 1.