TITLE: Are there finitely many sieves on each object in the distributive lattice cube category?
QUESTION [10 upvotes]: Define the distributive lattice cube category or Dedekind cube category $\square_{\land\lor}$ to be the full subcategory of the category of posets and monotone maps consisting of objects of the form $[1]^n \triangleq \{0 < 1\}^n$ for $n \ge 0$. Morphisms $[1]^m \to [1]^n$ in this category correspond to $n$-tuples of terms in $m$ variables in the theory of bounded distributive lattices, hence the name. (In terms of generators, every morphism can be written as a composite of face maps, degeneracies, permutations, diagonals, and min- and max-connections─this fact appears on the logical side as the disjunctive/conjunctive normal form presentations.)
The presheaf category $\mathrm{PSh}(\square_{\land\lor})$ is then one of many variations on "cubical sets". I'd like to know whether the following finiteness property is satisfied by the representable cubes in this category:
Question: In the presheaf category $\mathrm{PSh}(\square_{\land\lor})$, does every representable object have finitely many subobjects up to isomorphism? Equivalently, are there finitely many sieves on every object of $\square_{\land\lor}$?
In general, say that a category $\mathcal{C}$ has the finite sieve property if there are finitely many sieves on each object of $\mathcal{C}$. Recall that a sieve on an object $A$ in category $\mathcal{C}$ is a set $\mathcal{S}$ of arrows into $A$ such that if $g : B \to A$ is in $\mathcal{S}$ and $f : C \to B$ is an arbitrary arrow, then $g \circ f$ is in $\mathcal{S}$. To show that the set of sieves on some $A$ is finite, it suffices to show there are finitely many principal sieves, that is, sieves of the form $\langle g \rangle \triangleq \{g \circ f \mid f : C \to B\}$ for some $g : B \to A$. Using this and some specifics of $\square_{\land\lor}$, we can thus give the following more concrete reformulations:
Equivalent Question: Fix $k \ge 0$. Does there exist $\ell \ge 0$ such that every $f : [1]^n \to [1]^k$ generates the same principal sieve as some $f' : [1]^\ell \to [1]^k$?
Equivalent Question 2: Fix $k \ge 0$. Does there exist $\ell \ge 0$ such that for every $f : [1]^n \to [1]^k$, there are maps $[1]^n \overset{g}\to [1]^\ell \overset{h}\to [1]^n$ with $f \circ h \circ g = f$? (If this is so, then $\langle f \rangle = \langle f \circ h \rangle$.)
I'm interested in this cube category because of its relevance to modelling cubical type theories. In work with Christian Sattler we're in the process of writing up, we've found that finite product categories with the finite sieve property can be embedded in generalized Reedy categories in a useful way. We used this to characterize a certain model structure on $\square_\lor$ (the subcategory with just one connection, see remarks below), but don't know if any of the arguments apply with $\square_{\land\lor}$.

Some partial and related answers
The remainder of this post just collects special cases and related results that might or might not be helpful.
Remark: Any Eilenberg-Zilber category $\mathcal{C}$ in which there are finitely many face maps into any object has the finite sieve property.
Proof: Any $f : B \to A$ generates the same sieve as the degree-raising map of its Reedy factorization, as the degree-lowering map is a split epimorphism. $\blacksquare$
In particular, the simplex category and cartesian cube category (the wide subcategory of $\square_{\land\lor}$ generated by face maps, degeneracies, permutations, and diagonals) have the finite sieve property. If we add one connection to the latter, we get a category that is not generalized Reedy but still has the finite sieve property.
Proposition: The wide subcategory $\square_\lor$ of $\square_{\land\lor}$ consisting of maps generated by face maps, degeneracies, permutations, diagonals, and the max-connection $\lor$ has the finite sieve property. (Note: this is equivalently the wide subcategory of join-preserving maps.)
Proof: Let $f : [1]^n \to [1]^k$. Such a morphism corresponds to a tuple of $k$ terms in $n$ variables in the theory of bounded join-semilattices. We can assume without loss of generality that no component $f_j : [1]^n \to [1]$ for $1 \le j \le k$ is constant $1$. Then for each $f_j$, we have a normal form $f_j(x_1,\ldots,x_n) = \bigvee S_j$ for some $S_j \subseteq \{x_1,\ldots,x_n\}$. For $1 \le i \le n$, define $T_i = \{j \mid x_i \in S_j\} \subseteq \{1,\ldots,k\}$. If $n > 2^k$, then we must have $T_i = T_{i'}$ for some $i,i'$. Without loss of generality, say $T_1 = T_2$. Then $x_1$ and $x_2$ appear as disjuncts in the same components, so we have $f \circ d \circ c = f$ where $c : [1]^n \to [1]^{n-1}$ maps $(x_1,\ldots,x_n) \mapsto (x_1 \lor x_2, x_3, \ldots, x_n)$ and $d : [1]^{n-1} \to [1]^n$ maps $(y_1,\ldots,y_{n-1}) \mapsto (y_1,y_1,y_2, \ldots, y_{n-1})$. Thus $\langle f \rangle = \langle f \circ d \rangle$. By repeatedly reducing in this way, we find that $\langle f \rangle = \langle f' \rangle$ for some $f' : [1]^{2^k} \to [1]^k$. $\blacksquare$
Lemma: The finite sieve property holds for the objects $[1]^k$ with $k \le 2$ in $\square_{\land\lor}$.
Proof: The $k = 0$ case is obvious. For $k = 1$, any $f : [1]^n \to [1]$ is either constant $0$, constant $1$, or (we can see from the disjunctive normal form) satisfies $f \circ \Delta \circ f = f$ where $\Delta : [1] \to [1]^n$ is the diagonal $x \mapsto (x,\ldots,x)$.
Now $k = 2$, a bit sketchily. Suppose we have $f : [1]^n \to [1]^2$ and neither component is constant $0$ or $1$ (in which case we could reduce to the $k = 1$ case). We can write each component in disjunctive normal form, $f_j(x_1,\ldots,x_n) = \bigvee_{i \in I_j} \bigwedge C^j_i$ where $I_j$ is nonempty and each $C^j_i$ is a nonempty set of variables. If there exists some $C^1_i$ which is not a superset of any $C^2_{i'}$, choose one and call this $K_1$; otherwise choose an arbitrary $C^1_i$ to be $K_1$. (Note that in the latter case $f_1 \le f_2$.) Likewise, choose $K_2$ among the $C^2_i$ which is not a superset of any $C^1_{i'}$  if possible. Define $g : [1]^2 \to [1]^n$ by
$$
g(u_1,u_2)_i \triangleq \bigvee \{u_j \mid x_i \in K_j \}
$$
Then $f \circ g \circ f = f$. $\blacksquare$
In the $k \le 2$ case, we always get $f \circ g \circ f = f$ for some $g$. But this does not happen in general, as demonstrated by the following lower bound:
Proposition: For any $k \ge 1$, there exists some $f : [1]^n \to [1]^k$ such that $\langle f \rangle \neq \langle g \rangle$ for every $g : [1]^\ell \to [1]^k$ with ${\ell \choose {\lfloor \ell/2 \rfloor}} < 2^{k-1}$.
Proof: For any $v \in [1]^k$, write $\#_v \in [1]^{\lvert [1]^k \rvert}$ for the vector taking value $1$ in all coordinates except the $v$th. We define a monotone map $f : [1]^{\lvert [1]^k \rvert} \to [1]^k$.
$$
\begin{align*}
f(\top) &\triangleq \top \\
f(\#_v) &\triangleq v \\
f(\_) &\triangleq \bot &\text{otherwise}
\end{align*}
$$
Suppose we have $\langle f \rangle = \langle g \rangle$ for some $g : [1]^\ell \to [1]^k$, so there exist $h : [1]^n \to [1]^\ell$ and $t : [1]^\ell \to [1]^n$ with $f \circ t \circ h = f$ (and $g = f \circ t$). Then because $f^{-1}(v) = \{\#_v\}$ for $v \in [1]^k\setminus\{\bot,\top\}$, we must have $(t \circ h)(\#_v) = \#_v$ for these $v$. As $\#_v$ and $\#_w$ are incomparable for $v \neq w$, so must be $h(\#_v)$ and $h(\#_w)$. Thus $[1]^\ell$ must contain a set of at least $\lvert [1]^k\setminus\{\bot,\top\} \rvert = 2^{k-1}$ pairwise-incomparable vectors. The maximal antichain in $[1]^\ell$ has ${\ell \choose {\lfloor \ell/2 \rfloor}}$ elements, so ${\ell \choose {\lfloor \ell/2 \rfloor}} \ge 2^{k-1}$. $\blacksquare$
In particular, there is a principal sieve on $[1]^3$ not generated by any $f : [1]^3 \to [1]^3$.

REPLY [4 votes]: $\newcommand{\Pos}{\mathbf{Pos}}$
$\square_{\land\lor}$ does not have the finite sieve property. The following paper provided the key lemma:
Imrich, Wilfried; Kalinowski, Rafał; Lehner, Florian; Pilśniak, Monika, Endomorphism breaking in graphs, Electron. J. Comb. 21, No. 1, Research Paper P1.16, 13 p. (2014). ZBL1300.05100.
Write $C_n$ for the cycle graph on vertices $\{0,\ldots,n-1\}$. We can regard $C_{2n}$ as a bipartite graph, with
the even numbers in one part and the odd numbers in the other, and thereby as a poset with $i < j$ iff $i$ is
even, $j$ is odd, and $i-j = \pm 1 \bmod 2n$.
From the paper above, I just want the following (simple) fact, which is in their proof of Lemma 2.
Proposition 1. For any $n \ge 3$, there exists a 2-coloring of $C_{2n}$ such that the only color-preserving graph endomorphism of $C_{2n}$ is the identity function.
Corollary 2.
For any $n \ge 3$, there exists a poset map $f_n \colon C_{2n} \to C_4$ such that the only endomorphism of
$f_n \in \Pos/C_4$ is the identity function.
Proof. Choose a 2-coloring $c \colon C_{2n} \to \{0,1\}$ as in the proposition. Define $f \colon C_{2n} \to C_4$ by $f(i) \triangleq 2c(i) + i \mathbin{\%} 2$.
We have an embedding of posets $\iota \colon C_4 \hookrightarrow [1]^4$ defined as follows.
\begin{align*}
  0 &\mapsto 0001 & 1 &\mapsto 0111 \\
  2 &\mapsto 0010 & 3 &\mapsto 1011
\end{align*}
Given $i \in \mathbb{N}$, write $\chi_i \in [1]^n$ for the element with $1$ at index $i \bmod n$ and $0$ elsewhere. For any $n \ge 3$, we have a poset embedding $z_n \colon C_{2n} \hookrightarrow [1]^n$ defined as follows.
\begin{align*}
  z(i) &\triangleq \left\{
    \begin{array}{ll}
      \chi_{i/2} &\text{if $i$ is even} \\
      \chi_{\lfloor i/2 \rfloor} \vee \chi_{\lceil i/2 \rceil} &\text{if $i$ is odd}
    \end{array}
  \right.
\end{align*}
For each $n \ge 3$, we define a monotone map $g_n \colon [1]^n \to [1]^4$ by cases as follows.
\begin{align*}
  g_n(v) &\triangleq \left\{
    \begin{array}{ll}
      \iota(f_n(i)), &\text{if $\exists i < 2n.\ v = z_n(i)$} \\
      \bot, &\text{if $v = \bot$} \\
      \top, &\text{otherwise}
    \end{array}
  \right.
\end{align*}
By construction, we have $g_n \circ z_n = \iota \circ f_n$ as well as $g_n^{-1}(\iota(C_4)) = z_n(C_{2n})$.
Now, suppose we are given an endomorphism $h : g_n \to g_n$ in $\Pos/[1]^4$. Because
$h(z_n(C_{2n})) = h(g_n^{-1}(\iota(C_4))) \subseteq g_n^{-1}(\iota(C_4)) = z_n(C_{2n})$, there is a restriction
$h \upharpoonright z_n : C_{2n} \to C_{2n}$ fitting into the following commuting diagram in $\Pos$.
$\require{AMScd}$
\begin{CD}
C_{2n} @>{h \upharpoonright z_n}>> C_{2n} \\
@V{z_n}VV @V{z_n}VV \\
{[1]^n} @>h>> {[1]^n} \\
@V{g_n}VV @V{g_n}VV \\
{[1]^4} @= {[1]^4}
\end{CD}
It follows that we have $h \upharpoonright z_n \colon f_n \to f_n$ in $\Pos/C_4$, thus that
$h \upharpoonright z_n = \mathrm{id}$. Thus the image of any such $h \colon g_n \to g_n$ has cardinality at least $2n$; if
$h$ factors through some $[1]^\ell$, we must therefore have $\ell \ge \log_2 n + 1$.