Talk:F-algebra

message: The page claims that the natural numbers (together with [zero,succ]) are an initial algebra for Set and the functor FX = 1 + X. Is this really true? Surely there is an alegbra on an uncountably-infinite object...  —Preceding unsigned comment added by 158.130.51.58 (talk) 20:36, 31 May 2010 (UTC)
replies: text: Never mind, this is well explained on the (separate) page for initial algebras.  —Preceding unsigned comment added by 158.130.51.58 (talk) 20:53, 31 May 2010 (UTC)
thread_title:  Natural numbers are initial algebra? 

message: ::That was not immediately clear to me either, but then I could make sense of this notation. However, there remains another inaccuracy:
===A tuple of morphisms must have a product as the domain, but that's broken in the group example===
In F-algebra#Groups, the axiom:
:* ∀ x∈G, m(i(x), x) = m(x, i(x)) = e(*)
is rewritten as
:* m∘(i, Id) = m∘(Id, i) = e
as a formula or as an equivalent commutativity diagram:
(Now, we I'm readig the article, the diagram exactly corresponds to the formula, and they have the same problem, which is as follows.)
The domain of (Id, i) must actually be a product, namely, G×G (as seen in the other formulas with tuples of morphisms). (But it is assumed to be just G by the text in the article.)
Having the product as the domain misses the point that we should apply i to the same x as we pass as the other argument of m.
The formula needs a composition with the "duplicating" morphism dup from G to G×G. I don't know what a common notation for it would be. The axiom should be rewritten as:
:* m∘(i, Id)∘dup = m∘(Id, i)∘dup = e
Imz (talk) 16:57, 24 July 2019 (UTC)
replies: text: As of now, this is still a problem and it's still not fixed. I'd fix it, but I don't know how to make svg's for commuting diagrams. The fix itself is easy. See bottom of Help:Displaying a formula 67.198.37.16 (talk) 03:43, 18 March 2021 (UTC)
thread_title: Another problem: missing duplicating morphism

message: In F-algebra#Groups, the axiom:
:* ∀ x∈G, m(i(x), x) = m(x, i(x)) = e(*)
is rewritten as
:* m∘(i, Id) = m∘(Id, i) = e
as a formula or as an equivalent commutativity diagram:
(Now, we I'm readig the article, the diagram exactly corresponds to the formula, and they have the same problem, which is as follows.)
The domain of (Id, i) must actually be a product, namely, G×G (as seen in the other formulas with tuples of morphisms). (But it is assumed to be just G by the text in the article.)
Having the product as the domain misses the point that we should apply i to the same x as we pass as the other argument of m.
The formula needs a composition with the "duplicating" morphism dup from G to G×G. I don't know what a common notation for it would be. The axiom should be rewritten as:
:* m∘(i, Id)∘dup = m∘(Id, i)∘dup = e
Imz (talk) 16:57, 24 July 2019 (UTC)
replies: text: As of now, this is still a problem and it's still not fixed. I'd fix it, but I don't know how to make svg's for commuting diagrams. The fix itself is easy. See bottom of Help:Displaying a formula 67.198.37.16 (talk) 03:43, 18 March 2021 (UTC)
thread_title: A tuple of morphisms must have a product as the domain, but that's broken in the group example

message: There's another, related problem with that axiom--it's not well-typed. The text says the domain of e is the terminal object of the category, but the domain of m∘(i, Id) is G. Those two morphisms can't be equal.  — Preceding unsigned comment added by 2601:14F:4402:F1BC:C8A5:22CD:4375:261F (talk) 18:06, 21 December 2020 (UTC)
replies: text: It's actually e composed with the morphism to the terminal object.  So first map G to * then apply e. I don't know of what the "standard" notation for that is; its ... omitted by standard abuse of notation. Alas. 67.198.37.16 (talk) 05:00, 18 March 2021 (UTC)
thread_title: Not well typed

message: Most clearly, the following statement is wrong, but the way in which its wrong is illustrative of persistent errors in the first three example subsections:
:: Monoids are F-algebras of signature F(M) = 1 + M×M. In the same vein, semigroups are F-algebras of signature F(S) = S×S
This is simply incorrect; suppose we consider the functor F(S) = S×S defined from Set to Set. An F-algebra is just then a set with a binary operation - that is to say, it's a magma, not a semigroup; associativity is not encoded in that definition. Likewise, the things purporting to be monoids are really just magmas with an identified point, and the things purporting to be groups are magmas with an identified point and an identified unary operation - the axioms that are discussed are simply not encoded anywhere in the definition of an F-algebra. The page seems to make this misunderstanding explicit in statements like saying "with axioms to express associativity" when an F-algebra is simply not capable of expressing such a thing. It's true that semigroups are a full subcategory of the F-algebras over the given endofunctor, but I've never seen anyone use that fact or discuss algebraic structures in that language (though these things are good examples of algebras over a monad - but that's a slightly different concept and they're already used as examples on the page for monads).
I'm not so bold to make this edit without discussion (and being away from my books on category theory at the moment - can't check with them to see if I'm terribly mistaken or cite them), but I believe that the examples "Groups", "Algebraic Structures" and "Lattices" are all incorrect and should be deleted. (Perhaps expanding on the natural number example - it might be somewhat more illuminating to show how it's the initial F-algebra over F(X)=1+X and could set up other examples like lists and binary trees, which are already briefly mentioned in the article)
Meeloq (talk) 02:18, 24 April 2021 (UTC)
replies: text: I think it's more accurate to say that the examples are incomplete and misleading. I can see the point that they are driving at, but there do seem to be some steps that involve sleight-of-hand and prestidigitation. It would be best to fix up these sections with an appropriate reference in one's lap. This kind of abstraction is relatively common, I've seen it more than a few times, in a number of different textbooks, its a provides a nice way of defining groups and other algebraic structures without having to pose the axioms in terms of specific group elements.  Just the way it's explained here remains awkward and obtuse and unsatisfying. 67.198.37.16 (talk) 18:14, 26 May 2021 (UTC)
thread_title:  Most of the examples look to be incorrect 

message: In the groups section there were previously 3 laws:
:* m\circ(m, \mathrm{Id}) = m\circ(\mathrm{Id}, m)
:* m\circ(e, \mathrm{Id}) = m\circ(\mathrm{Id}, e) = \mathrm{Id}
:* m\circ(i, \mathrm{Id}) = m\circ(\mathrm{Id}, i) = e
Now in the first two laws we have this notation where if f : A \rightarrow B and g : C \rightarrow D, then (f,g) : A \times C \rightarrow B \times D.
However in the third law there is a subtle issue.  Here the same notation is used to represent a different idea: if f : A \rightarrow B and g : A \rightarrow C, then (f,g) : A \rightarrow B \times C.  Now abuse of notation is common in math, and if we try to use this interpretation on the first two laws the types don't work out, so the interpretation is clearly incorrect.
But the issue is that if we assume that the notation being used in the third law is the same, which seems like the normal thing to do, the types do work out.  We get an interpretation of the law which is a algebraic law but is entirely incorrect.  The law states \forall x y, m(x,i(y))=e(*), which is not the proper law but is a law.
I think this makes it a rather poor way to present the topic.
It's basically a trap of notation.
I don't know how to make this better, but the commutative diagrams do not have this issue.  Since the types are labeled we know that (\mathrm{Id},m) : G \rightarrow G\times G and not (\mathrm{Id},m) : G\times G \rightarrow G\times G, so we get forced into an interpretation by the types.
For this reason I've removed the confusing expression all together and left the commutative diagrams to fill the space.
If someone knows some way to restate the laws to avoid the confusing ambiguity present please do so.
AquitaneHungerForce (talk) 14:52, 1 January 2022 (UTC)
thread_title:  Removed laws from the Groups section.