Are all monomorphisms in the category of bounded lattices regular?

Question

Let $\mathbf{Lat}_{01}$ be the category of bounded lattices with lattice homomorphisms that respect the smallest and the largest element. Is there a monomorphism in $\mathbf{Lat}_{01}$ that is not regular?

Answer 1

Is there a monomorphism in $\mathbf{𝐋𝐚𝐭}_{01}$ that is not regular?

No, all monomorphisms in the variety $\mathbf{𝐋𝐚𝐭}_{01}$ are regular.

Reasoning: The variety $\mathbf{𝐋𝐚𝐭}_{01}$ has the strong amalgamation property. This fact may be found online in the table near the bottom of this page, but the usual citation for this is to:

Jónsson, Bjarni
Universal relational systems.
Math. Scand. 4 (1956), 193-208.

Here is the relevance of the strong amalgamation property (SAP). Suppose that $f\colon L\to K$ is a monomorphism. Since $\mathbf{𝐋𝐚𝐭}_{01}$ is a variety, monomorphisms are injective, so we may (and do) view $L$ as a $01$-sublattice of $K$ and view $f$ as inclusion. To prove that $f$ is regular, we need a bounded lattice $J$ and morphisms $h,k\colon K\to J$ such that $L$ is the equalizer sublattice of $h$ and $k$. This means that $h$ and $k$ agree on the sublattice $L$ and on no other element of $K$. To find $J, h, k$, just apply strong amalgamation to the span $K\stackrel{f}{\leftarrow} L \stackrel{f}{\rightarrow} K$ to get $J\in \mathbf{𝐋𝐚𝐭}_{01}$ and morphisms $h, k\colon K\to J$ which, by SAP, agree on the sublattice $L$ and nowhere else. This realizes $L$ as an equalizer sublattice, hence the inclusion morphism $f$ is regular.

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Edit Sep 2, 2024.

I am adding more to my answer to respond to a comment made earlier today by Jochen Wengenroth.

In my original answer, I wrote that the statement that the variety $\mathbf{Lat}_{01}$ has the strong amalgamation property may be found online at Peter Jipsen's math structures wiki, but that the usual citation for this is to Jónsson's 1956 paper. Jochen has stated that there are no references at Jipsen's page. This is partly true: there are some references under the Acknowledgments link on Jipsen's wiki, most importantly to the reference:

E. W. Kiss, L. Márki, P. Pröhle, W. Tholen,
Categorical algebraic properties. A compendium on amalgamation, congruence extension, epimorphisms, residual smallness, and injectivity,
Studia Sci. Math. Hungar., 18, 1982, 79-140, 85k:18003

which Jipsen mentions in the first line of this page. This paper by Kiss-Marki-Prohle-Tholen in turn cites Jónsson's 1956 paper. Let me mention also that the paper by Kiss-Marki-Prohle-Tholen cites a paper of Max Kelley from 1969 for a result implying that SAP for a variety implies that monomorphisms are regular. (See Proposition 6.1.)

Regarding my statement that the usual citation for the fact that $\mathbf{Lat}_{01}$ is to Jónnson's 1956 paper, let me justify this by mentioning some relevant works that cite it:

Alan Day and Jaroslav Jezek
The Amalgamation Property for Varieties of Lattices
Transactions of the American Mathematical Society, Volume 286, Number 1, (1984), 251-256.

This paper proves that the only varieties of lattices with the amalgamation property are the trivial variety, the variety of distributive lattices, and the variety of all lattices. The attribution to Jónnson for the fact that $\mathbf{Lat}$ has SAP is made in the first line of the introduction.

Also see on pages 135-137 of the book

Peter Jipsen and Henry Rose
Varieties of lattices
Springer, Lecture Notes in Mathematics 1533, 1992, 176 pages

for the attribution to Jónnson of the fact that $\mathbf{Lat}$ has SAP. Jipsen and Rose give a short, straightforward proof of SAP for $\mathbf{Lat}$ (not necessarily bounded lattices) in Theorem 6.18 on page 137.

The proof of SAP given in Jipsen and Rose is for the variety $\mathbf{Lat}$ rather than $\mathbf{Lat}_{01}$, but it is easy to see that $\mathbf{Lat}$ has the SAP iff $\mathbf{Lat}_{01}$ has the SAP. Let me explain this now.

Assume that $\mathbf{Lat}$ has the SAP. To derive that $\mathbf{Lat}_{01}$ also has the SAP:

1. Assume that $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ is a span in $\mathbf{Lat}_{01}$ with $f, g$ embeddings. Let's also call this an SAP amalgam; our goal is to complete it in $\mathbf{Lat}_{01}$.
2. Ignoring the constants, $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ is an amalgam in $\mathbf{Lat}$.
3. Use SAP in $\mathbf{Lat}$ to obtain embeddings $f'\colon B\to D$ and $g'\colon C\to D$ such that $f′\circ f=g′\circ g$ and $\textrm{im}(f')\cap\textrm{im}(g')=\textrm{im}(f′\circ f)$. (I.e., find an SAP solution to the amalgam in $\mathbf{Lat}$.)
4. Observe that $f(0_A)=0_B$ and $g(0_A)=0_C$, since $f$ and $g$ were originally $\mathbf{Lat}_{01}$ morphisms, so we have equalities in $D$: $$f'(0_B)=f'\circ f(0_A) = g'\circ g(0_A) = g'(0_C).$$ Call this common element $m\in D$. Similarly, $$f'(1_B)=f'\circ f(1_A) = g'\circ g(1_A) = g'(1_C).$$ Call this common element $M\in D$.
5. The images of $f'\colon B\to D$ and $g'\colon C\to D$ are contained entirely in the interval $[m,M]_D$ of $D$, and $f'$ and $g'$ maps the bottoms and tops of their domains to $m$ and $M$ respectively. Replace $D$ with this interval sublattice $D'=[m,M]_D$ and add constants $0_{D'}, 1_{D'}$ interpreted so that $0_{D'} = m$ and $1_{D'} = M$. With these interpretations, $D'$ become an object of $\textbf{Lat}_{01}$ and the same functions $f'\colon B\to D'$ and $g'\colon C\to D'$ are embeddings in $\mathbf{Lat}_{01}$ which yield an SAP solution to the amalgam $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ in $\mathbf{Lat}_{01}$. This shows that if $\mathbf{Lat}$ has SAP, then $\mathbf{Lat}_{01}$ also has SAP.

Now for the converse. Assume that $\mathbf{Lat}_{01}$ has the SAP.

1. There is a functor that assigns to each lattice $L\in \mathbf{Lat}$ a lattice $L_0^1\in \mathbf{Lat}_{01}$ obtained by adjoining a new top element $1$ and new bottom element $0$. $\mathbf{Lat}$ morphisms $f\colon L\to K$ extend uniquely to $\mathbf{Lat}_{01}$ morphisms $f_0^1\colon L_0^1\to K_0^1$.
2. Each amalgam $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ in $\mathbf{Lat}$ extends uniquely to $B_0^1\stackrel{f_0^1}{\leftarrow} A_0^1 \stackrel{g_0^1}{\rightarrow} C_0^1$ in $\mathbf{Lat}_{01}$. Complete it in $\mathbf{Lat}_{01}$ with, say, $f'\colon B_0^1\to D$, $g'\colon C_0^1\to D$.
3. Ignoring constants, $f'|_B\colon B\to D$, $g'|_C\colon C\to D$ is an SAP solution to $B\stackrel{f}{\leftarrow} A \stackrel{g}{\rightarrow} C$ in $\mathbf{Lat}$.

Let me close this addendum by addressing Jochen's question. Q: The question is whether the morphisms constructed in POS making $𝑓$ an equalizer are lattice morphisms. A: See Lemma 6.17 of Jipsen-Rose for a proof of this.

Answer 2

This is a concrete version of Keith Kearnes's answer:

Let $A=f(X)$ be the image of a lattice monomorphism $f:X\to Y$ and consider in $Y\times\{0,1\}$ the partial order defined by $(x,i)\le (y,i)$ if $x\le y$ and for different $i,j\in\{0,1\}$ take $(x,i)\le (y,j)$ if there is $a\in A$ with $x\le a$ and $a\le y$. Now, consider the quotient $Z$ of $Y\times\{0,1\}$ which identifies points $(a,0)$ and $(a,1)$ for $a\in A$. Let $q$ be the corresponding quotient map $q:Y\times\{0,1\}\to Z$ and $g_i:Y\to Z$, $y\mapsto q(y,i)$.

$Z$ is a partially ordered set (maybe not a lattice because there are new incomparable $q(y,0)$ and $q(y,1)$ for $y\in Y\setminus A$) and $g_i$ preserve finite meets and joins: To see that, e.g., $q(x\vee y,0)=q(x,0) \vee q(y,0)$, one has to consider new upper bounds $q(z,1)$ of $\{q(x,0),q(y,0)\}$ with $z\notin A$. By definition of the partial order, there are $a,b\in A$ with $x\le a\le z$ and $y\le b\le z$ so that $A\ni a\vee b\le z$ which implies $q(x\vee y,0)\le q(a\vee b,0)=q(a\vee b,1)\le q(z,1)$.

Finally, let $i$ be the order embedding of $Z$ into its Dedekind-MacNeill completion $\hat Z$ which is a lattice (even a complete one). The embedding $i$ preserves meets and joins which exist in $Z$ and therefore, $i\circ g_j:Y\to \hat Z$ are lattice morphisms with $$\{y\in Y:i\circ g_0(y)=i\circ g_1(y)\}=\{y\in Y:g_0(y)=g_1(y)\}=A.$$ This implies that $f$ is a regular monomorphism.

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Edit. Essentially the same arguments show that also in the category of complete lattices and meet and join preserving maps, all monomorphisms are regular.