Explicit lifting characterization of complete lattices among posets?

Question

It's well-known that the complete lattices are characterized among all posets as the regular-injectives. That is, a poset $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to the class $Emb$ of all embeddings of posets. This remarkable characterization would be even more useful if there were some smaller, more explicit class of embeddings $\mathcal E \subset Emb$ sufficient to check injectivity. That is,

Questions:

1. Is there a good sub-class $\mathcal E \subset Emb$ of all poset embeddings such that a poset $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to all the embeddings in $\mathcal E$?

2. What if we restrict attention to finite posets? That is: Is there a good subclass $\mathcal E^{fin} \subset Emb^{fin}$ of all embeddings of finite posets such that a finite poset $L$ is a lattice if and only if $L$ has the right lifting property with respect to all the embeddings in $\mathcal E^{fin}$?

What is meant by "good" is a bit subjective, but to start I'd be happy with anything non-tautological. Ultimately, it would be nice to have a class $\mathcal E$ or $\mathcal E^{fin}$ which is "explicit" in some sense, so that it actually becomes easier to check that a poset $L$ is complete via lifting properties. Normally, I'd hope for $\mathcal E$ to be small, but I'm pretty sure this is not possible.

One candidate I have in mind for $\mathcal E$ would be the collection of embeddings $\{S \to S^\triangleright \mid S \text{ discrete}\} \cup \{S \to S^\triangleleft \mid S \text{ discrete}\}$ which add a new top or bottom element to a discrete poset $S$. But I suspect this is too naive, as it would mean that any $\infty$-directed and $\infty$-codirected poset is a complete lattice -- this is probably false.

Answer 1

You are right that $\mathcal E$ cannot be small because complete lattices are not closed in posets under $\lambda$-filtered colimits for any regular cardinal $\lambda$. A candidate for $\mathcal E$ consists from embeddings of posets to their Mac-Neille completions. Since the Mac-Neille completion of a finite poset is finite, it works in the finite case too. But I think that you would not consider it to be "good".

Answer 2

After a bit of thought, I think I have a pretty good set for $\mathcal E^{fin}$ and a pretty good class for $\mathcal E$.

Proposition: Let $L$ be a finite poset. Then $L$ is a lattice if and only if $L$ has the right lifting property with respect to the following set $\mathcal E^{fin}$ of embeddings:

1. $\emptyset \to 1$

2. $2 \to 2^\triangleleft$ and $2 \to 2^\triangleright$ where $2$ is the discrete poset with two elements and these morphisms add a cone and cocone respectively.

3. $\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie \to \bbowtie$ where $\bbowtie$ is the 5-element poset $x,y < a < p,q$ and $\bowtie$ is the full sub-poset on $x,y,p,q$.

Proof: It suffices to show that $L$ is a meet-semilattice, i.e. that for every finite set $S \subseteq L$, the poset $L \downarrow S$ of elements under $S$ has a top element. It suffices to consider the cases (i) when $S$ is empty and (ii) when $S$ has two elements. Moreover, every finite directed poset has a top element, so it will suffice to show that (i) $L$ is directed and (ii) for every $p, q \in L$, the poset $L \downarrow \{p,q\}$ is directed. (i) follows from (1) and the second part of (2). (ii) follows from the first part of (2) and (3).

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Analogously, using $\infty$-directedness in place of directedness, we have

Proposition: Let $L$ be a poset. Then $L$ is a complete lattice if and only if $L$ has the right lifting property with respect to the following class $\mathcal E$ of embeddings:

1. $\emptyset \to 1$

2. $S \to S^\triangleleft$ and $S \to S^\triangleright$ for each discrete poset $S$ (i.e. we add a top element and a bottom element, respectively, to $S$)

3. $\newcommand{\bbowtie}{\bowtie\mkern-17mu\bullet\mkern17mu}$ $\bowtie_{S,S} \to \bbowtie_{S,S}$ for each set $S$, where $\bbowtie_{S,T}$ is the poset whose under lying set is $S \amalg T \amalg \{a\}$, with $S < a < T$, and $\bowtie_{S,T}$ is the full subposet on $S \amalg T$.