Mixing
Does the category $mathsf{DL}$ of bounded distributive lattices have (filtered) colimits?
How do I show that the category $mathsf{DL}$ of bounded distributive lattices and monotone maps has (filtered) colimits? I guess we have an adjunction
$$
F dashv U : mathsf{DL} to mathsf{Set}.
$$
Then $U$ preserves limits and $F$ preserves colimits. Can we deduce from this that $mathsf{DL}$ has (filtered) colimits, and if so, how?
category-theory
asked Nov 19 '18 at 23:35
Math Student 020
946616
add a comment |
How do I show that the category $mathsf{DL}$ of bounded distributive lattices and monotone maps has (filtered) colimits? I guess we have an adjunction
$$
F dashv U : mathsf{DL} to mathsf{Set}.
$$
Then $U$ preserves limits and $F$ preserves colimits. Can we deduce from this that $mathsf{DL}$ has (filtered) colimits, and if so, how?
category-theory
asked Nov 19 '18 at 23:35
Math Student 020
946616
add a comment |
How do I show that the category $mathsf{DL}$ of bounded distributive lattices and monotone maps has (filtered) colimits? I guess we have an adjunction
$$
F dashv U : mathsf{DL} to mathsf{Set}.
$$
Then $U$ preserves limits and $F$ preserves colimits. Can we deduce from this that $mathsf{DL}$ has (filtered) colimits, and if so, how?
category-theory
asked Nov 19 '18 at 23:35
Math Student 020
946616
How do I show that the category $mathsf{DL}$ of bounded distributive lattices and monotone maps has (filtered) colimits? I guess we have an adjunction
$$
F dashv U : mathsf{DL} to mathsf{Set}.
$$
Then $U$ preserves limits and $F$ preserves colimits. Can we deduce from this that $mathsf{DL}$ has (filtered) colimits, and if so, how?
category-theory
category-theory
asked Nov 19 '18 at 23:35
Math Student 020
946616
asked Nov 19 '18 at 23:35
Math Student 020
946616
asked Nov 19 '18 at 23:35
Math Student 020
946616
asked Nov 19 '18 at 23:35
Math Student 020
946616
asked Nov 19 '18 at 23:35
Math Student 020
946616
946616
add a comment |
add a comment |
1 Answer
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oldest
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The below assumes that your morphisms of bounded distributive lattices are maps preserving meets, joins, top, and bottom.
Your adjunction, like any free-forgetful adjunction between categories of models of an algebraic theory, is monadic. In particular, it creates reflexive coequalizers. This tells us, essentially, that we can construct unique bounded distributive lattice (bdl) structures on the quotient set of a bdl by a congruence: an equivalence relation closed under the lattice operations. Now, the usual construction of colimits from coproducts and coequalizers actually uses only reflexive coequalizers. Thus to give all colimits of bdls it suffices to give their coproducts. And this is easy, in terms of generators and relations: a presentation of the coproduct of a family of bdls is just the disjoint union of presentations of each bdl in the family. Of course, as with free products of groups, the actual elements of such a coproduct can be tricky to get a handle on. But it certainly exists!
Filtered colimits are even easier: the forgetful functor, again, creates them. This isn't a property of general monads, but of those corresponding to algebraic theories in particular. The point is that the filtered colimits of the underlying sets of some filtered family $L_i$ of bdls has, again, a unique bdl structure making the inclusion maps into homomorphisms. It's given simply by $[a_i]vee [b_i]=[a_ivee b_i]$, $0=[0_i]$ etc. The filteredness serves to show that these formulae define all possible meets and joins in a well defined way and that the resulting operations are distributive.
All of the above holds for the category of models of any finitary algebraic theory, which covers most familiar categories of algebra with exceptions like fields (axioms can only be equations), posets (there are only operations, no relations), and complete lattices (operations must be of finite arity.)
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
add a comment |
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The below assumes that your morphisms of bounded distributive lattices are maps preserving meets, joins, top, and bottom.
Your adjunction, like any free-forgetful adjunction between categories of models of an algebraic theory, is monadic. In particular, it creates reflexive coequalizers. This tells us, essentially, that we can construct unique bounded distributive lattice (bdl) structures on the quotient set of a bdl by a congruence: an equivalence relation closed under the lattice operations. Now, the usual construction of colimits from coproducts and coequalizers actually uses only reflexive coequalizers. Thus to give all colimits of bdls it suffices to give their coproducts. And this is easy, in terms of generators and relations: a presentation of the coproduct of a family of bdls is just the disjoint union of presentations of each bdl in the family. Of course, as with free products of groups, the actual elements of such a coproduct can be tricky to get a handle on. But it certainly exists!
Filtered colimits are even easier: the forgetful functor, again, creates them. This isn't a property of general monads, but of those corresponding to algebraic theories in particular. The point is that the filtered colimits of the underlying sets of some filtered family $L_i$ of bdls has, again, a unique bdl structure making the inclusion maps into homomorphisms. It's given simply by $[a_i]vee [b_i]=[a_ivee b_i]$, $0=[0_i]$ etc. The filteredness serves to show that these formulae define all possible meets and joins in a well defined way and that the resulting operations are distributive.
All of the above holds for the category of models of any finitary algebraic theory, which covers most familiar categories of algebra with exceptions like fields (axioms can only be equations), posets (there are only operations, no relations), and complete lattices (operations must be of finite arity.)
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
add a comment |
The below assumes that your morphisms of bounded distributive lattices are maps preserving meets, joins, top, and bottom.
Your adjunction, like any free-forgetful adjunction between categories of models of an algebraic theory, is monadic. In particular, it creates reflexive coequalizers. This tells us, essentially, that we can construct unique bounded distributive lattice (bdl) structures on the quotient set of a bdl by a congruence: an equivalence relation closed under the lattice operations. Now, the usual construction of colimits from coproducts and coequalizers actually uses only reflexive coequalizers. Thus to give all colimits of bdls it suffices to give their coproducts. And this is easy, in terms of generators and relations: a presentation of the coproduct of a family of bdls is just the disjoint union of presentations of each bdl in the family. Of course, as with free products of groups, the actual elements of such a coproduct can be tricky to get a handle on. But it certainly exists!
Filtered colimits are even easier: the forgetful functor, again, creates them. This isn't a property of general monads, but of those corresponding to algebraic theories in particular. The point is that the filtered colimits of the underlying sets of some filtered family $L_i$ of bdls has, again, a unique bdl structure making the inclusion maps into homomorphisms. It's given simply by $[a_i]vee [b_i]=[a_ivee b_i]$, $0=[0_i]$ etc. The filteredness serves to show that these formulae define all possible meets and joins in a well defined way and that the resulting operations are distributive.
All of the above holds for the category of models of any finitary algebraic theory, which covers most familiar categories of algebra with exceptions like fields (axioms can only be equations), posets (there are only operations, no relations), and complete lattices (operations must be of finite arity.)
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
add a comment |
The below assumes that your morphisms of bounded distributive lattices are maps preserving meets, joins, top, and bottom.
Your adjunction, like any free-forgetful adjunction between categories of models of an algebraic theory, is monadic. In particular, it creates reflexive coequalizers. This tells us, essentially, that we can construct unique bounded distributive lattice (bdl) structures on the quotient set of a bdl by a congruence: an equivalence relation closed under the lattice operations. Now, the usual construction of colimits from coproducts and coequalizers actually uses only reflexive coequalizers. Thus to give all colimits of bdls it suffices to give their coproducts. And this is easy, in terms of generators and relations: a presentation of the coproduct of a family of bdls is just the disjoint union of presentations of each bdl in the family. Of course, as with free products of groups, the actual elements of such a coproduct can be tricky to get a handle on. But it certainly exists!
Filtered colimits are even easier: the forgetful functor, again, creates them. This isn't a property of general monads, but of those corresponding to algebraic theories in particular. The point is that the filtered colimits of the underlying sets of some filtered family $L_i$ of bdls has, again, a unique bdl structure making the inclusion maps into homomorphisms. It's given simply by $[a_i]vee [b_i]=[a_ivee b_i]$, $0=[0_i]$ etc. The filteredness serves to show that these formulae define all possible meets and joins in a well defined way and that the resulting operations are distributive.
All of the above holds for the category of models of any finitary algebraic theory, which covers most familiar categories of algebra with exceptions like fields (axioms can only be equations), posets (there are only operations, no relations), and complete lattices (operations must be of finite arity.)
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
The below assumes that your morphisms of bounded distributive lattices are maps preserving meets, joins, top, and bottom.
Your adjunction, like any free-forgetful adjunction between categories of models of an algebraic theory, is monadic. In particular, it creates reflexive coequalizers. This tells us, essentially, that we can construct unique bounded distributive lattice (bdl) structures on the quotient set of a bdl by a congruence: an equivalence relation closed under the lattice operations. Now, the usual construction of colimits from coproducts and coequalizers actually uses only reflexive coequalizers. Thus to give all colimits of bdls it suffices to give their coproducts. And this is easy, in terms of generators and relations: a presentation of the coproduct of a family of bdls is just the disjoint union of presentations of each bdl in the family. Of course, as with free products of groups, the actual elements of such a coproduct can be tricky to get a handle on. But it certainly exists!
Filtered colimits are even easier: the forgetful functor, again, creates them. This isn't a property of general monads, but of those corresponding to algebraic theories in particular. The point is that the filtered colimits of the underlying sets of some filtered family $L_i$ of bdls has, again, a unique bdl structure making the inclusion maps into homomorphisms. It's given simply by $[a_i]vee [b_i]=[a_ivee b_i]$, $0=[0_i]$ etc. The filteredness serves to show that these formulae define all possible meets and joins in a well defined way and that the resulting operations are distributive.
All of the above holds for the category of models of any finitary algebraic theory, which covers most familiar categories of algebra with exceptions like fields (axioms can only be equations), posets (there are only operations, no relations), and complete lattices (operations must be of finite arity.)
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
answered Nov 20 '18 at 1:27
Kevin Carlson
32.5k23271
32.5k23271
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
add a comment |
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
1
1
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
This is very helpful, thanks!
– Math Student 020
Nov 20 '18 at 1:53
add a comment |
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