 powerset lattices?

Let $S$ be a finite set, and let $2^S$ be its powerset, regarded as a lattice. Let $L$ be a quotient (in the category of lattices and maps which preserve $\top,\bot,\wedge,\vee$) of $S$. What can we say about $L$?
 In fact, what I'd really like to know is: which finite semilattices are retracts (via $\bot,\vee$-preserving maps) of finite powerset lattices. I think the two questions are equivalent via a short-but-not-immediate argument.
Note that if $P$ is an arbitrary finite poset, then the lattice $2^P$ of poset maps $P \to 2$ is an example of such a lattice.

A:

The class of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are exactly the lattice reducts of finite Boolean algebras.
In particular, any quotient $L$ of $2^S$ has to be a bounded distributive lattice, as the class of distributive lattices is a variety. Moreover, if $x\in L$ is an image of a set $A\subseteq S$, then the image of $S\smallsetminus A$ is an element $y$ such that $x\lor y=\top$ and $x\land y=\bot$. Thus, $L$ is complemented, i.e., it is a Boolean algebra, and as such it is isomorphic to $2^{S'}$ for some set $S'$.

A:

Tim-
Emil answered the question you asked, but since you wrote ``what I'd really like to know is: which finite semilattices are retracts (via $\bot$, $\vee$-preserving maps) of finite powerset lattices'' let me add to his answer.
The $2$-element semilattice is injective in the class of semilattices. 
The class of injectives is closed under products and retracts.
Up to isomorphism, the powers of the $2$-element semilattice are the power-set semilattices. Hence retracts of power-set semilattices must be injective.
Conversely, since the $2$-element semilattice is the only subdirectly irreducible semilattice, every semilattice is embeddable in some power
powerset lattices? 

 Let $ S$ be a finite set, and let $ 2^S$ be its powerset, involve as a wicket. Let $ L$ be a quotient (in the category of lattice and function which preserve $ \top,\bot,\wedge,\vee$) of $ S$. What can we say approximately $ L$? 
  In fact, what I'd really like to know is: which finite semilattices are retracts (via $ \bot,\vee$-preserving map) of finite powerset lattices. I think the two doubt are equivalent via a short - but - not - immediate argumentation. 
 notice that if $ P$ is an arbitrary finite poset, then the lattice $ 2^P$ of poset maps $ phosphorus \to 2 $ is an example of such a wicket. 

 A: 

 The course of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are precisely the lattice reducts of finite Boolean algebras. 
 In particular, any quotient $ L$ of $ 2^S$ have to be a bounded distributive lattice, as the class of distributive lattices is a variety. Moreover, if $ x\in L$ is an picture of a set $ A\subseteq S$, then the image of $ S\smallsetminus A$ is an element $ y$ such that $ x\lor y=\top$ and $ x\land y=\bot$. Thus, $ L$ is complemented, i.e., it is a Boolean algebra, and as such it is isomorphous to $ 2^{S'}$ for some fit $ S'$. 

 A: 

 Tim- 
 Emil answered the question you asked, but since you wrote '` what I'd really like to know is: which finite semilattices are retracts (via $ \bot$, $ \vee$-preserving maps) of finite powerset lattice" let me add to his solution. 
 The $ 2$-element semilattice is injective in the course of semilattices. 
 The class of injectives is closed under products and retracts. 
 Up to isomorphism, the powers of the $ 2$-element semilattice are the office - jell semilattices. Hence retracts of power - jell semilattices must be injective. 
 Conversely, since the $ 2$-element semilattice is the only subdirectly irreducible semilattice, every semilattice is embeddable in some power
 poaerset lattices?

Let $S$ be x finite set, anb let $2^S$ be ita powersdt, regarded as a lattice. Let $L$ be a quotient (in the categury of lantices ane maks which preserve $\top,\bot,\wcbge,\ves$) of $V$. What can we ssy about $L$?
 In fact, what H'd rzally like to know is: which finite sqmilattocfs are retractf (vis $\bot,\bve$-kreserving maps) of finite powersef lattibes. I think the teo questions are equivalenh viw a short-but-not-immfdiate argunent.
Gite that if $O$ is an arbitrary finije poset, then the lattice $2^P$ of puset kaps $P \to 2$ iw aj example of such a lattice.

A:

Tmv class of finote powerset lsttmces is closed under quotments, up to isomorphysm. That hs, the quotients arw wxactny tve lxrtize devucfs of vinmte Boolean algebras.
In particular, any quouiegn $L$ of $2^S$ has fo be w founded distributive lattice, as the clavs kf distributive latticew is a variety. Moreovgr, if $x\in J$ is an image of a set $A\subseteq S$, then the image of $S\aoalosctmivys A$ is an element $y$ such that $x\lor y=\top$ and $x\lwhd y=\not$. Thus, $L$ is coiplemented, o.e., iy is a Boolean algebxz, znd as such it is lsomorpric ti $2^{S'}$ for sjme xet $S'$.

A:

Tim-
Emil answered the wuestion you qsked, but since yob wrote ``what I'd teally like to know is: which yinite semilatticfs are reffacts (via $\bot$, $\ved$-prvsereing maps) of finite powersqt latticxs'' lec me add to nis anfwer.
The $2$-elfment semilattice is injfctivg in tve class ov semilattices. 
The class of injxrtives is cloxeg utder probucts snd retracts.
U[ to isomorphixm, the 'owers of the $2$-elvment semmlattice are the power-set demilattices. Hence rqtraxts if powef-ret semilatticrs must bv nnjective.
Xonversely, since tme $2$-elgmsnt semilattice is the only subditecgly igrevucibja semilattica, evdry xemilxttice is enceddsble in some power
powerset lattices? Let $S$ be a finite let be its regarded as a quotient the category of and maps which $\top,\bot,\wedge,\vee$) of $S$. What can we about $L$? In fact, what I'd really like to know is: which finite are retracts (via $\bot,\vee$-preserving maps) of finite powerset lattices. I think the two are via short-but-not-immediate Note that if $P$ is an arbitrary finite poset, then the lattice $2^P$ of poset maps \to 2$ is an example of such a A: The class of powerset lattices is closed under up isomorphism. That the are the lattice reducts finite Boolean algebras. In particular, any quotient $L$ of $2^S$ has to be a bounded distributive lattice, the class lattices is variety. if L$ is an a set $A\subseteq S$, then the A$ is an element $y$ such that $x\lor and $x\land Thus, $L$ is complemented, i.e., it a Boolean algebra, and as such it is to $2^{S'}$ for some set $S'$. A: Tim- Emil answered the question you asked, but wrote ``what I'd really to know is: finite are (via $\vee$-preserving maps) finite powerset lattices'' let me add to his answer. The $2$-element is injective in the class of semilattices. The class of closed products and retracts. to isomorphism, the powers the semilattice are the power-set retracts power-set injective. since $2$-element semilattice is the subdirectly irreducible semilattice, every semilattice embeddable in some power
 powerset lattices?

Let $S$ be a fiNite set, and Let $2^S$ bE itS poWeRset, RegaRded as a lattice. lEt $L$ bE a quotient (in the categorY of laTtICes aND mAps whIch presERvE $\TOp,\bOt,\WeDge,\VeE$) Of $s$. What Can We say abOut $L$?
 In fact, WhaT I'D really like tO KnOw is: which fIniTe semilatticEs aRe retrAcTs (vIA $\bot,\vEe$-pReserVing maPS) of finIte powersEt LAtticeS. i think tHE TwO queStions are equivaleNT vIA a short-but-not-iMmediaTe ARgUMEnt.
notE that if $P$ is An ArbitRAry finiTE pOSET, thEN the lattice $2^P$ oF poset maps $P \TO 2$ is An examPlE of SUch a laTtice.

a:

THE clAss of finite PoweRset lattiCes is cLOsed undER quotieNts, up tO isOmoRphiSM. THaT is, ThE QuoTIeNts ARe eXactly thE lAtTice rEducTS OF FiniTe BOoleAn algEbras.
In particUlaR, any QUotIent $L$ Of $2^S$ haS to bE a BoundEd distRibutIvE lattice, as the clAss oF distribuTivE lAttIcEs is a VAriety. morEovEr, if $x\in l$ is an imAGe oF a SET $a\sUbseteq S$, then the imaGe OF $s\sMallsetmInus A$ iS An ElEMent $y$ sucH tHat $X\lor Y=\TOp$ and $X\lanD Y=\bOt$. Thus, $L$ iS complEMeNtEd, i.e., it iS a booleaN aLgeBra, And as SUch iT is isoMorphic tO $2^{S'}$ for SOme set $S'$.

A:

Tim-
EmiL Answered the quEStION yOU askEd, bUt since you wRote ``WHat I'D reaLLy LikE To knoW is: whIcH FiNIte semilattices are rEtRacts (vIa $\bot$, $\Vee$-preserving Maps) of finiTE POwerset lAttiCEs'' LEt me add to his anSwer.
THe $2$-element sEMilatticE is inJective iN the class OF SemilattIceS. 
ThE clAss OF InJectives is cloSED undEr ProductS anD retracTs.
UP to IsoMorPhIsm, the powErs of the $2$-ElEmEnT sEmiLattiCE are the pOwEr-sEt SemIlattICes. HenCe retRactS oF pOWer-Set semiLAtTICes mUsT bE injEctIvE.
ConvErseLY, siNce the $2$-eLement semIlaTTice Is ThE only suBdirectly irreDuCible semilAtTicE, every SEMilatticE is embeddable in some poweR
 powerset lattices?

Let $ S$ be a fi niteset , a nd  let  $2^ S$ be its powe r set,  regarded as a lattice . Let  $ L $ be aquoti ent (in th e cat eg or y o fl at tices  an d mapswhich pres erv e$\top,\bot,\ w ed ge,\vee$)of$S$. What ca n w e sayab out $L$? In  fact , what I'd re ally like  t o  knowi s: whic h fi nite  semilattices are re t racts (via $\b ot,\ve e$ - pr e s erv ing  maps) offi nitep owerset la t t i ces .  I think thetwo questio n s a re equ iv ale n t viaa sho rt - but -not-immedi ateargument. 
Notet hat if$ P$ is a n arbi tra ryfini t epo set ,t hen th e l a tti ce $2^P$  o fposet  map s $ P  \to  2$  isan ex ample of such  alatt i ce. 

A:
Theclas sof fi nite p owers et  lattices is cl osed  under qu oti en ts,  u p toi somorp his m.That is , the q u oti en t s ar e exactly the latt ic e re ducts of  finit e  B oo l ean alge br as. 
Inp a rticu lar, an y quotie nt $L$ of  $ 2^S$ ha sto beabou nde d dis t ribu tive l attice,as th e  class of dist r ibutive latti c es i sa  var iet y. Moreover , if $x\i n L$ is  an image  of a  s e t$ A\subseteq S$, then  t he ima ge of  $S\smallsetm inus A$ is a n  element  $y$ su c h that $x\lory=\to p$ and $x\ l and y=\b ot$.Thus, $L $ is comp l e mented,i.e .,itisa Bo olean algebra , andas  such i t i s isomo rph icto$2^ {S '}$ for s ome set$S '$ .
A :
Tim-E mil answ er edth e q uesti o n youasked , bu tsi n ceyou wro t e` ` what  I 'd  rea lly  l ike t o kn o w i s: whic h finitesem i latt ic es  are re tracts (via $ \b ot$, $\vee $- pre servin g maps) of  finite powerset lattic e s'' let  me  addto h is answer .
T he $2$ -el e ment s emilat ticeis  in j e ctive i nthe  c lass of se m i lat tices .
The  classof injectives is c l ose d under produ cts  and r et rac t s. Upto iso m o rphism, the pow ers of the  $ 2 $- element se m ila tt ice are  the po wer-s e t semil attices.Hence ret ra ctso f  po wer-set se milattic es must b e  inje c ti ve.
C onv ersely ,sin ce th e $2$- e lem ent s emilat ti ce isthe o nl y subdir ectly irreducible semil attice , eve rysemilatti cei s e mbeddable  insome power 
 powerset_lattices?

Let $S$_be a finite set,_and let_$2^S$_be its_powerset,_regarded as a_lattice. Let $L$_be a quotient (in_the category of_lattices_and maps which preserve $\top,\bot,\wedge,\vee$) of $S$. What can we say about $L$?
 In_fact,_what I'd_really_like_to know is: which finite_semilattices are retracts (via $\bot,\vee$-preserving_maps) of_finite powerset lattices. I think the two questions_are_equivalent via a_short-but-not-immediate argument.
Note that if $P$ is an arbitrary finite_poset, then the lattice $2^P$ of_poset maps $P_\to_2$_is an example of_such a lattice.

A:

The class of finite_powerset lattices is closed under quotients,_up to isomorphism. That is, the quotients_are exactly the lattice reducts of_finite Boolean algebras.
In particular, any_quotient $L$_of $2^S$ has to be_a bounded distributive_lattice, as_the class of_distributive lattices is a variety. Moreover,_if $x\in L$_is an image of a set_$A\subseteq_S$, then the_image_of_$S\smallsetminus A$_is an element_$y$_such that_$x\lor_y=\top$ and $x\land y=\bot$. Thus, $L$_is_complemented, i.e., it is a Boolean algebra,_and as such it_is_isomorphic to $2^{S'}$ for_some set $S'$.

A:

Tim-
Emil answered the_question you asked, but since you_wrote ``what_I'd really_like to know is: which finite semilattices are retracts (via $\bot$,_$\vee$-preserving maps) of finite powerset lattices''_let me add to_his answer.
The_$2$-element_semilattice is injective_in_the class_of semilattices. 
The class of injectives is_closed under_products and retracts.
Up to isomorphism, the_powers of the $2$-element_semilattice_are the power-set semilattices. Hence retracts_of power-set semilattices must be injective.
Conversely,_since the $2$-element semilattice is_the_only_subdirectly irreducible semilattice, every semilattice_is embeddable in some power