Injection between posets $(\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)$ and $(\{-, 0,+\}^d,\preceq)$, where $-\preceq 0$ and $+\preceq 0$.

Question

Let $(\{-, 0,+\}, \preceq)$ be the poset defined by the relations $-\preceq 0$ and $+\preceq 0$ and let $\left(\{-, 0,+\}^d, \preceq\right)$ be its $d$ -fold product. Show that: $$ (\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)\cong(\{-, 0,+\}^d,\preceq), $$ where $\Phi([-1,1]^d)$ is the face lattice of the $d$-dimensional cube.

I have an idea of what the isomorphism would be, but I cannot yet prove that it’s injective.

First, a remark is that, since the product of posets is a binary operation, the elements of $(\{-, 0,+\}^d, \preceq)$ technically have a lot of extra parentheses, which could be omitted without losing any important information. So, for example, instead of $((0,+),-)$, we’d simply have $(0,+,-)$.

Let $F \in \Phi([-1,1]^d)$ denote a face of the cube, and in particular let $v$ denotes a vertex. Then the mapping I’m thinking of is defined as:

\begin{align} &\phi:(\Phi([-1,1]^d)\setminus\{\emptyset\},\subseteq)\to(\{-, 0,+\}^d,\preceq) \\ &\phi(F) = (\text{sgn}(\sum_{v_i \in F}v_{i1}),…,\text{sgn}(\sum_{v_i \in F}v_{id})) \end{align} i.e. we look at the vertices contained in $F$, summing up their entries component-wise, and then taking their signs.

Now I want to show that this map is injective. Given two faces $F_1$ and $F_2$ such that $\phi(F_1) = \phi(F_2)$, it’s sufficient to show that a vertex $v_i = (v_{i1},…,v_{id})$ in $F_1$ is also in $F_2$ (and vice versa). Consider an entry $j$ in $\phi(F_1) = \phi(F_2) \in \{-, 0,+\}^d$. If $F_1$ and $F_2$ contain just one vertex, then this entry determines exactly the sum at position $j$ of the vertices in $F_1$ and $F_2$, respectively (namely, the “sum” of just one summand, resulting in value $-1$ or $1$).

If $F_1$ and $F_2$ both contain $2^m$ vertices, then we can continue in an inductive fashion to show that the summands that result in the $j$ entry of $\phi(F_1)$ and $\phi(F_2)$ match each other one by one: in summing the $m$ entries in the $j$-th position of vertices in $F_1$ and those in $F_2$, assuming that the partial sums up to a certain position $n < m$ match, then the $j$ entry of $\phi(F_1)$ and $\phi(F_2)$ determines what the value of the next summands must be.

Of course, this only works if $F_1$ and $F_2$ have the same number of vertices. I got stuck here since I can’t show that given $\phi(F_1) = \phi(F_2)$, it cannot be the case that $F_1$ contains $m$ vertices, and $F_2$ contains $m’$ vertices, where $m \neq m’$.

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Edit: I’ve found another map, which might be easier to check for the properties of an isomorphism. It’s not yet entirely rigorous, but I’m just putting it here for comments.

For $F \in \Phi([-1,1]^d), \ \phi(F) = (u_1,…u_d) := u \in (\{-, 0,+\}^d, \preceq)$, where $u$ satisfies the following:
$\cdot$ $u$ has |dim($F$)| $0$’s, at positions corresponding to the axes of $\mathbb{R}^d$ that $F$ is “parallel” to.
$\cdot$ The other positions in $u$ correspond to the remaining axes in $\mathbb{R}^d$, have values either $1$ or $-1$, depending whether $F$ lies in the positive or negative side of those axes, respectively.

Answer 1

First, let me clarify some notation:

• The poset $(\{-, 0,+\}^d, \preceq)$ to me means the set of d-tuples $(u_1,\dots,u_d)$ with each component $u_i\in \{-,0,+\}$, and with $u\preceq u'$ iff $u_i\preceq u_i'$ for all $i$.
• The face lattice of the d-dimensional cube, $\Phi([-1,1]^d)$, can be expressed as follows: A face of a d-dimensional cube has the form $F=f_1 \times f_2 \times \cdots \times f_d$, where each $f_i$ is either the singleton $\{-1\}$, the singleton $\{+1\}$, or the interval $[-1,1]$. Then $\Phi([-1,1]^d)$ is the set of all such faces, together with the empty set $\emptyset$ (required to make it a lattice), and the order is usual set inclusion $\subseteq$. In particular, $f_1\times\cdots\times f_d \subseteq f_1'\times\cdots\times f_d'$ iff $f_i \subseteq f_i'$ for all $i$.

Note that both $(\{-, 0,+\}^d, \preceq)$ and $\Phi([-1,1]^d) \setminus \emptyset$ contain $3^d$ elements, so to show they are isomorphic it suffices to find an injection which preserves the order.

The isomorphism is essentially what the OP suggested in the edit. Define $\psi:(\{-, 0,+\}^d, \preceq) \rightarrow \Phi([-1,1]^d) \setminus \emptyset $ by $\psi(u_1,\dots,u_d) = f_1\times \cdots \times f_d$ where $f_i = \{\pm 1\}$ if $u_i=\pm$ and $f_i = [-1,1]$ if $u_i = 0$. This is clearly a bijection, so it just remains to show that it preserves the order.

For $u,u'\in \{-, 0,+\}^d$ we have

$$ \begin{align*} u\preceq u' & \iff \forall (i) \;\; u_i \preceq u_i' \\ & \iff \forall(i)\;\; \psi(u)_i \subseteq \psi(u')_i \\ & \iff \psi(u) \subseteq \psi(u') \end{align*} $$ where the second line is the only non-trivial one, and follows immediately from our definition of $\psi$ and the properties of the order $\preceq$. Hence $\psi$ preserves the order, and we are done.