Mixing
Is a poset an algebra?
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
I encountered some nice material about so-called universal algebras.
E.g. a lattice is an algebra (in the sense of universal algebra), and can be presented as a tuple $langle L,wedge,veerangle$ where $wedge$ and $vee$ are binary operations on $L$.
I am not (yet) familiar with that stuff, and one of the first questions that arose was:
Can a partial order be presented as an algebra?
I think not because I cannot really find operations on it with some (finite) arity.
Am I correct in my thinking?
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
category-theory universal-algebra
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
asked Aug 31 at 14:35
Vera
1,818413
add a comment |Â
up vote
9
down vote
favorite
I encountered some nice material about so-called universal algebras.
E.g. a lattice is an algebra (in the sense of universal algebra), and can be presented as a tuple $langle L,wedge,veerangle$ where $wedge$ and $vee$ are binary operations on $L$.
I am not (yet) familiar with that stuff, and one of the first questions that arose was:
Can a partial order be presented as an algebra?
I think not because I cannot really find operations on it with some (finite) arity.
Am I correct in my thinking?
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
category-theory universal-algebra
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
asked Aug 31 at 14:35
Vera
1,818413
3
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I encountered some nice material about so-called universal algebras.
E.g. a lattice is an algebra (in the sense of universal algebra), and can be presented as a tuple $langle L,wedge,veerangle$ where $wedge$ and $vee$ are binary operations on $L$.
I am not (yet) familiar with that stuff, and one of the first questions that arose was:
Can a partial order be presented as an algebra?
I think not because I cannot really find operations on it with some (finite) arity.
Am I correct in my thinking?
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
category-theory universal-algebra
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
asked Aug 31 at 14:35
Vera
1,818413
I encountered some nice material about so-called universal algebras.
E.g. a lattice is an algebra (in the sense of universal algebra), and can be presented as a tuple $langle L,wedge,veerangle$ where $wedge$ and $vee$ are binary operations on $L$.
I am not (yet) familiar with that stuff, and one of the first questions that arose was:
Can a partial order be presented as an algebra?
I think not because I cannot really find operations on it with some (finite) arity.
Am I correct in my thinking?
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
category-theory universal-algebra
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
asked Aug 31 at 14:35
Vera
1,818413
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
edited Sep 2 at 16:14
Alex Kruckman
23.7k22453
23.7k22453
asked Aug 31 at 14:35
Vera
1,818413
asked Aug 31 at 14:35
Vera
1,818413
asked Aug 31 at 14:35
Vera
1,818413
1,818413
3
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58
add a comment |Â
3
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58
3
3
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.
In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,leq)$, one defines an algebra $(A,cdot)$, where
$$
a cdot b =
begincases
a &textif a leq b,\
b &textotherwise.
endcases
$$
The poset can be recovered from the algebra by making $a leq b$ iff $acdot b = a$.
In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,leq, 1)$ with a top element they define two algebraic structures:
$$
a to b =
begincases
1 &textif a leq b,\
b &textotherwise.
endcases
$$
and
$$
a cdot b =
begincases
b &textif a leq b,\
1 &textotherwise.
endcases
$$
These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.
answered Aug 31 at 19:08
amrsa
3,3432518
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
add a comment |Â
up vote
10
down vote
No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=aleq b,cleq d$ with the equivalence relation $sim$ generated by $bsim c$. The quotient by this equivalence relation is the poset $Q=aleq [b]leq d$, where by transitivity $aleq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:Pto Q$ to $p^-1(aleq d)to aleq d$. Now, $p^-1(aleq d)$ is just the discrete poset $a,d$. But the morphism $a,dtoaleq d$ is not the quotient map for the restriction of $sim$ to $a,d$. Indeed, it's not the quotient by any equivalence relation at all.
This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $sim$ and we restrict the quotient map $Ato A/sim$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $Gto G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $gsim hiff gh^-1in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $sim$, so the restriction of the quotient map is indeed a quotient map.
To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $Xtimes X$ such that the diagonal map $Xto Xtimes X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $Xtimes Xto Xtimes X$ that interchanges the factors (symmetry), and the pullback of $Rtimes_X Rsubset Xtimes Xtimes X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.
So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
 |Â
show 1 more comment
up vote
3
down vote
From the question:
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
There is a theorem of algebra that says
Theorem. A bijective morphism is an isomorphism.
That is, if $varphi: Ato B$ is a homomorphism that is 1-1 and onto, then $varphi^-1$ is also a homomorphism. More categorically, the theorem says that, if $mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.
This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.
You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.
In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,leq)$, one defines an algebra $(A,cdot)$, where
$$
a cdot b =
begincases
a &textif a leq b,\
b &textotherwise.
endcases
$$
The poset can be recovered from the algebra by making $a leq b$ iff $acdot b = a$.
In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,leq, 1)$ with a top element they define two algebraic structures:
$$
a to b =
begincases
1 &textif a leq b,\
b &textotherwise.
endcases
$$
and
$$
a cdot b =
begincases
b &textif a leq b,\
1 &textotherwise.
endcases
$$
These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.
answered Aug 31 at 19:08
amrsa
3,3432518
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
add a comment |Â
up vote
5
down vote
accepted
A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.
In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,leq)$, one defines an algebra $(A,cdot)$, where
$$
a cdot b =
begincases
a &textif a leq b,\
b &textotherwise.
endcases
$$
The poset can be recovered from the algebra by making $a leq b$ iff $acdot b = a$.
In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,leq, 1)$ with a top element they define two algebraic structures:
$$
a to b =
begincases
1 &textif a leq b,\
b &textotherwise.
endcases
$$
and
$$
a cdot b =
begincases
b &textif a leq b,\
1 &textotherwise.
endcases
$$
These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.
answered Aug 31 at 19:08
amrsa
3,3432518
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.
In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,leq)$, one defines an algebra $(A,cdot)$, where
$$
a cdot b =
begincases
a &textif a leq b,\
b &textotherwise.
endcases
$$
The poset can be recovered from the algebra by making $a leq b$ iff $acdot b = a$.
In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,leq, 1)$ with a top element they define two algebraic structures:
$$
a to b =
begincases
1 &textif a leq b,\
b &textotherwise.
endcases
$$
and
$$
a cdot b =
begincases
b &textif a leq b,\
1 &textotherwise.
endcases
$$
These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.
answered Aug 31 at 19:08
amrsa
3,3432518
A poset can indeed be given an algebraic structure.
This is not a generalization of a lattice, but it's an algebra, nevertheless.
I suppose there's a plethora of ways of doing this, but I'll just refer three of them, in which two only apply to posets with a maximum element.
In the paper The variety generated by order algebras, by Ralph Freese, Jaroslav Jezek, Peter Jipsen, Petar Markovic, Miklos Maróti and Ralph McKenzie, published in Algebra Universalis 47 (2002), given a poset $(P,leq)$, one defines an algebra $(A,cdot)$, where
$$
a cdot b =
begincases
a &textif a leq b,\
b &textotherwise.
endcases
$$
The poset can be recovered from the algebra by making $a leq b$ iff $acdot b = a$.
In Algebras defined from ordered sets and the varieties they generate, by Joel Berman and Wilhelm Blok, published in Order 23 (2006), for posets $(P,leq, 1)$ with a top element they define two algebraic structures:
$$
a to b =
begincases
1 &textif a leq b,\
b &textotherwise.
endcases
$$
and
$$
a cdot b =
begincases
b &textif a leq b,\
1 &textotherwise.
endcases
$$
These are both very nice papers I read some years ago.
I don't recall all the details in them, but I'm just mentioning that these studies (and perhaps others) have been done, and these definitions of algebras exist, in the first case, for any poset at all.
answered Aug 31 at 19:08
amrsa
3,3432518
answered Aug 31 at 19:08
amrsa
3,3432518
answered Aug 31 at 19:08
amrsa
3,3432518
answered Aug 31 at 19:08
amrsa
3,3432518
3,3432518
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
add a comment |Â
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
2
2
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
There's also The algebraic theory of order. This doesn't produce a variety but does produce an essentially algebraic theory. In this case, the trick is to use the functor sending a poset to its underlying relation rather than the carrier set.
– Derek Elkins
Aug 31 at 19:26
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
This is actually enough already as answer to my question. The tag "category-theory" that I used somehow placed my question in a broader perspective that I was not aware of. Thank you!
– Vera
Sep 1 at 10:51
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@Vera I recommend exploring the categorical perspective. Category theory has a very good story for universal algebra. It also reveals a quite useful, hidden generality of universal algebraic reasoning. As an analogy, when one first learns about linear algebra, usually it's over the field of real numbers. However, nearly every theorem and proof of the first few years of linear algebra would work unchanged for arbitrary fields (or at least fields of characteristic 0). Universal algebra works just as well if instead of carrier sets, we have carrier topological spaces or carrier manifolds.
– Derek Elkins
Sep 2 at 1:10
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
@DerekElkins That indeed sounds quite promising. I am glad with this answer (which meets exactly my (first) needs), but also with the one of Kevin, and will surely go deeper into the categorical perspective (as you mention). Thank you for your encouragement.
– Vera
Sep 2 at 15:03
add a comment |Â
up vote
10
down vote
No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=aleq b,cleq d$ with the equivalence relation $sim$ generated by $bsim c$. The quotient by this equivalence relation is the poset $Q=aleq [b]leq d$, where by transitivity $aleq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:Pto Q$ to $p^-1(aleq d)to aleq d$. Now, $p^-1(aleq d)$ is just the discrete poset $a,d$. But the morphism $a,dtoaleq d$ is not the quotient map for the restriction of $sim$ to $a,d$. Indeed, it's not the quotient by any equivalence relation at all.
This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $sim$ and we restrict the quotient map $Ato A/sim$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $Gto G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $gsim hiff gh^-1in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $sim$, so the restriction of the quotient map is indeed a quotient map.
To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $Xtimes X$ such that the diagonal map $Xto Xtimes X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $Xtimes Xto Xtimes X$ that interchanges the factors (symmetry), and the pullback of $Rtimes_X Rsubset Xtimes Xtimes X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.
So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
 |Â
show 1 more comment
up vote
10
down vote
No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=aleq b,cleq d$ with the equivalence relation $sim$ generated by $bsim c$. The quotient by this equivalence relation is the poset $Q=aleq [b]leq d$, where by transitivity $aleq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:Pto Q$ to $p^-1(aleq d)to aleq d$. Now, $p^-1(aleq d)$ is just the discrete poset $a,d$. But the morphism $a,dtoaleq d$ is not the quotient map for the restriction of $sim$ to $a,d$. Indeed, it's not the quotient by any equivalence relation at all.
This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $sim$ and we restrict the quotient map $Ato A/sim$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $Gto G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $gsim hiff gh^-1in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $sim$, so the restriction of the quotient map is indeed a quotient map.
To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $Xtimes X$ such that the diagonal map $Xto Xtimes X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $Xtimes Xto Xtimes X$ that interchanges the factors (symmetry), and the pullback of $Rtimes_X Rsubset Xtimes Xtimes X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.
So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
 |Â
show 1 more comment
up vote
10
down vote
up vote
10
down vote
No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=aleq b,cleq d$ with the equivalence relation $sim$ generated by $bsim c$. The quotient by this equivalence relation is the poset $Q=aleq [b]leq d$, where by transitivity $aleq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:Pto Q$ to $p^-1(aleq d)to aleq d$. Now, $p^-1(aleq d)$ is just the discrete poset $a,d$. But the morphism $a,dtoaleq d$ is not the quotient map for the restriction of $sim$ to $a,d$. Indeed, it's not the quotient by any equivalence relation at all.
This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $sim$ and we restrict the quotient map $Ato A/sim$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $Gto G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $gsim hiff gh^-1in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $sim$, so the restriction of the quotient map is indeed a quotient map.
To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $Xtimes X$ such that the diagonal map $Xto Xtimes X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $Xtimes Xto Xtimes X$ that interchanges the factors (symmetry), and the pullback of $Rtimes_X Rsubset Xtimes Xtimes X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.
So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
No, posets are not algebras, at least in the most common interpretation of that term. The problem is with quotients of equivalence relations. Consider the poset $P=aleq b,cleq d$ with the equivalence relation $sim$ generated by $bsim c$. The quotient by this equivalence relation is the poset $Q=aleq [b]leq d$, where by transitivity $aleq d$ is forced. It's this extra "generated" relation that causes the trouble. Consider the restriction of the map $p:Pto Q$ to $p^-1(aleq d)to aleq d$. Now, $p^-1(aleq d)$ is just the discrete poset $a,d$. But the morphism $a,dtoaleq d$ is not the quotient map for the restriction of $sim$ to $a,d$. Indeed, it's not the quotient by any equivalence relation at all.
This kind of phenomenon doesn't happen in categories of algebras. If $A$ is an algebra for an algebraic theory with an equivalence relation $sim$ and we restrict the quotient map $Ato A/sim$ to some subalgebra of the codomain, we always get a map which is also a quotient by an equivalence relation. The explanation for this is that quotients by equivalence relations are constructed in algebras by just taking the quotient on the underlying sets, then canonically inducing new operations. In the most familiar example of groups, this story corresponds to thinking of a quotient map $Gto G/N$ by a normal subgroup instead as being the quotient by the equivalence relation $gsim hiff gh^-1in N$. (In most algebraic theories, including some familiar ones like monoids, there's no useful notion of normal subobject, so the equivalence relation approach is much more general.) A subgroup of $G/N$ has its inverse image a subgroup of $G$ closed under $sim$, so the restriction of the quotient map is indeed a quotient map.
To see that this argument proves that it's impossible to realize posets as algebras for an algebraic theory, we should clarify a few things. First, none of the concepts I've used above depend on the choice of forgetful functor into sets. For instance, by an "equivalence relation" on an object $X$ of a category with finite limits I mean a subobject $R$ of $Xtimes X$ such that the diagonal map $Xto Xtimes X$ factors through $R$ (reflexivity), $R$ factors through its composite with the map $Xtimes Xto Xtimes X$ that interchanges the factors (symmetry), and the pullback of $Rtimes_X Rsubset Xtimes Xtimes X$ over the first and last factors of $X$ factors through $R$ (transitivity.) The "quotient by an equivalence relation" is a regular epimorphism.
So, categorically, the argument has been that in the category of posets, the pullback of a regular epimorphism need not be a regular epimorphism, unlike in any category of algebras. This is one of the defining properties of a regular category, so the reason that posets are not algebras is that they don't form a regular category.
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
edited Aug 31 at 18:13
Derek Elkins
15.2k11035
15.2k11035
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
answered Aug 31 at 17:46
Kevin Carlson
29.5k23066
29.5k23066
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
 |Â
show 1 more comment
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
1
1
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
I was going to ask "how do you know that the relation you consider is a congruence for the given algebra ?", but the last two paragraphs clarified that
– Max
Aug 31 at 18:00
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
@Max Great, sorry for the somewhat mysterious sequencing.
– Kevin Carlson
Aug 31 at 18:21
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
Thank you. I will need some time to digest this and will have a closer look at regular categories. You speak of "quotients of equivalence relations". Don't you mean the stronger concept "congruences" (instead of "equivalence relations") here? Another question: in the material that I quote in my question algebra's are always - if I am right - sets $A$ that are accompanied by some operations on $A$ (like multiplication, et cetera). I cannot find any operations on a poset. Do they indeed lack, and - if so - isthat also a legal argument for saying that posets are not algebra's?
– Vera
Aug 31 at 18:59
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
The only thing I can think of is some boolean that states that $aleq b$ is true or is not true, but that is not a function $A^2to A$ (so no operation) but is a function $A^2to0,1$.
– Vera
Aug 31 at 19:01
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
About your last question: Kevin's answer shows that there is no possible set of operations to put on posets that would make them algebras: there is no category of algebras and algebra morphisms that is equivalent to the category of posets and monotone maps
– Max
Aug 31 at 19:36
 |Â
show 1 more comment
up vote
3
down vote
From the question:
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
There is a theorem of algebra that says
Theorem. A bijective morphism is an isomorphism.
That is, if $varphi: Ato B$ is a homomorphism that is 1-1 and onto, then $varphi^-1$ is also a homomorphism. More categorically, the theorem says that, if $mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.
This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.
You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
add a comment |Â
up vote
3
down vote
From the question:
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
There is a theorem of algebra that says
Theorem. A bijective morphism is an isomorphism.
That is, if $varphi: Ato B$ is a homomorphism that is 1-1 and onto, then $varphi^-1$ is also a homomorphism. More categorically, the theorem says that, if $mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.
This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.
You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
add a comment |Â
up vote
3
down vote
up vote
3
down vote
From the question:
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
There is a theorem of algebra that says
Theorem. A bijective morphism is an isomorphism.
That is, if $varphi: Ato B$ is a homomorphism that is 1-1 and onto, then $varphi^-1$ is also a homomorphism. More categorically, the theorem says that, if $mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.
This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.
You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
From the question:
More generally: if I see the objects of some category as candidate for being an algebra of some type, then are there criteria that can be applied in order to check this?
There is a theorem of algebra that says
Theorem. A bijective morphism is an isomorphism.
That is, if $varphi: Ato B$ is a homomorphism that is 1-1 and onto, then $varphi^-1$ is also a homomorphism. More categorically, the theorem says that, if $mathcal C$ is any full subcategory of the category of all algebras of some signature, then the forgetful functor to Sets reflects isomorphisms.
This theorem states a simple necessary condition for a concrete category to be equivalent to a concrete category of algebras. To apply it to the category of posets, observe that there is an order-preserving bijection between the 2-element discrete poset (=totally unordered poset) and the 2-element chain. Its inverse is not order preserving.
You can easily apply this criterion to many other concrete categories, like the categories of graphs, topological spaces, ETC.
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
answered Sep 3 at 18:15
Keith Kearnes
5,2091626
5,2091626
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
add a comment |Â
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
Your last statement is in error; this criterion fails in many concrete categories and the category of topological spaces is a counterexample. For example, the map $(cos theta, sin theta)$ from $[0, 2 pi)$ to the unit circle is a bijective morphism that is not an isomorphism. Similarly, for any topological space $X$, let $|X|$ be the discrete space consisting of the same points. Then $|X| to X$ is a bijective morphism that is not an isomorphism (unless $X$ is discrete).
– Hurkyl
Sep 3 at 18:21
3
3
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
You are misreading my post, which explains that a category FAILING the criterion is not algebraic. It was my claim that the category of topological spaces FAILS the criterion and is therefore not algebraic. (Notice from the 2nd-to-last paragraph that by "apply the criterion" I mean "show that your category has a bijective morphism that is not an isomorphism".)
– Keith Kearnes
Sep 3 at 18:50
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
So if I understand well then you provide a suitable necessary condition on a category for being algebraic, right? Thank you. The comments on this question show that I was not too much busy with categories, but I am also glad answers of this sort.
– Vera
Sep 4 at 11:43
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2900769%2fis-a-poset-an-algebra%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Do you just want to know if every poset can be seen as an algebra, or if the category of posets can be seen as the category of algebras for some signature?
– Arnaud D.
Aug 31 at 15:07
@ArnaudD. Both actually, but I must confess that I do not really see any difference between the two interpretations of my question. (If posets are algebra's then the category of posets has objects that are algebra's...). Can you explain please? Is it about morphisms in that category maybe?
– Vera
Aug 31 at 19:06
Yes, it's about the morphisms. If you want, you can read my comment as "Do you also want order-preserving maps between posets to coincide with homomorphisms between the corresponding algebras?". Note that Kevin's answer assumes that.
– Arnaud D.
Aug 31 at 19:38
@ArnaudD. I understand. No I had no morphisms in mind when I posted the question. So the tag "category-theory" was maybe not fully justified. On the other hand I have no regrets that I placed that tag. The answer of Kevin is very welcome anyhow.
– Vera
Sep 1 at 10:58