Talk:F-algebra

1 + X[edit]

What does "1 + X" in the example mean? I (not a category theorist) assume it's the set of numbers y such that y = 1 + x for some x in X. But that only makes sense if X is a set of numbers (or I guess any ring with 1). That doesn't seem to extend very well to the entire category of sets. Staecker 02:11, 25 July 2006 (UTC)[reply]

I'm not a category theorist either, but I'm guessing it refers to the disjoint union. That is, X + Y is the union of the sets X and Y, but with the elements "relabelled" to make sure the sets don't overlap. See category of sets and coproduct to see how this idea fits into category theory.

I believe the article is wrong when it says "where 1 is some singleton subset of X". If we are indeed talking about the disjoint union then 1 can refer to any singleton set. GilesEdkins 00:22, 21 January 2007 (UTC)[reply]

I've clarified the notations of the example. I hope that they are clear now. Samuel Mimram 17:56, 22 January 2007 (UTC)[reply]

Thanks- should've known it was a coproduct. Staecker 19:32, 22 January 2007 (UTC)[reply]

Natural numbers are initial algebra?[edit]

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)[reply]

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)[reply]

Technical?[edit]

Diego Moya has done some nice cleaning up of this article. But he added the tag Template:Technical. I think this is inherently a technical topic; I don't think someone should really hope to understand much if they don't know what a category is. Could you make some suggestions about what you think is TOO technical, where it doesn't need to be? Similarly on F-coalgebra. Thanks, ComputScientist (talk) 19:35, 21 January 2011 (UTC)[reply]

When an article is technical in nature, it still can be made accessible to a general public by placing it in context. Create an applications section like that in F-coalgebra#Applications showing examples applied to programming, explain who use this technique and what for, tell something about its history (who invented it, for what reasons)...

I don't buy the typical argument often found at mathematics talk pages that "this article surely would be useless to someone who doesn't know X". Every article has a human side even if the topic is extremely technical or abstract. An encyclopedia should aim to cover that side; to learn only about the technical details you'd use a textbook instead. The point is not that someone would understand 'much'; is that the reader could at least understand something. If someone (as myself) who doesn't know cathegory theory found this article, she must have some good reason to want to learn about it. An article written by and for mathematicians experts in the topic is useless for that reader.

(Also the technical tag was added when the lead section was a mathematical definition, something absolutely contrary to wp:lead. The changes I propose are to continue further in the direction I began at the introduction). Diego Moya (talk) 12:37, 22 January 2011 (UTC)[reply]

Notation: what is the meaning of a composition of a binary operation with a tuple?[edit]

This statement is completely opaque to me: m∘(m, Id) = m∘(Id, m). If this is not a composition, then what is this? Wvxvw (talk) 15:36, 6 May 2015 (UTC)[reply]

I agree that this is not really clear. To clarify, given a category C which has all finite products, f from A to B and g from X to Y, you can define a morphism (f,g) from A×X to B×Y.

Thus in our case, if m is a morphism from A×A to A, then (m, Id) is a morphism from (A×A)×A to A×A, so composition with m remains possible to form m∘(m, Id) as a morphism from (A×A)×A to A.

Sedrikov (talk) —Preceding undated comment added 19:18, 29 December 2015 (UTC)[reply]

Another problem: missing duplicating morphism[edit]

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[edit]

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)[reply]

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)[reply]

Not well typed[edit]

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)[reply]

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)[reply]

Groups as categories[edit]

The morphisms of the category (G,m,e,i) are presumably a subset of the C-automorphisms of G (since there is really nothing else to work with). But not all C-automorphism of G will do, only those which preserve the structure provided by m,e, and i. That is: the f in Morph_C(G,G) that are group homomorphisms with respect to the multiplication m, identity e, and inverse i. Or:

     m°(f,f) = f°m : G x G -> G
     f°e = e°f = e : G -> G
     f°i = i°f : G -> G

One cannot conclude from these requirements on f that f is invertible: as a counterexample, let f be the morphism e composed with the unique map to the terminal object (G -> * -> G). So invertibility is an additional assumption, but why? Invertibility of "(invisible) elements of G" is provided by i, and invertibility of the automorphisms of G is not necessary (and not true).

It feels as if we are confusing "the elements of the group G" with the automorphisms of G...

_________________

The 2nd group axiom reads ∀ x∈G, ∃ 1∈G, m(1, x) = m(x, 1) = x. Should that not rather be ∃ 1∈G, ∀ x∈G, m(1, x) = m(x, 1) = x. Otherwise the neutral element could be different for every x. Skolemization would also give e(x) and not e(*). Is it admissible to quantify over the neutral element in the two axioms separately. I doubt it, but have no counterexample at hand (Mattias Ulbrich (talk) 16:09, 9 August 2017 (UTC))[reply]

_________________

It is also hard to follow the flow of the reasoning.

"The group G.." -- there is no antecedent to justify the implication of uniqueness from "the."

"The group G is an object in C" -- so some randomly chosen group is an object in an arbitrarily chosen category ...?

Do we mean

"A group, G, may be defined by requiring that there be morphisms in C m: GxG -> G, e:*->G, and i:G -> G such that ..."?

or

"A group, G, exists when there are morphisms in C m: GxG -> G, e:*->G, and i:G -> G such that ..."?

My problems continue as I read on to understand what has been said thusfar.

The claim is the F(G) = 1+G+GxG is a functor. In order to define an F-algebra it must be an endofunctor, F:C->C for some category C. What is C?

If C = {G,m,e,i} then the image F(G) can only be the only object in {G,m,e,i}, i.e. G, and thus G = 1+G+GxG?

If C = Groups then

why is the set direct sum specified rather than the Group direct sum (if there is one)?
what is the group structure on 1+G+GxG?

If C != {G,m,e,i}

why did we spend all this time developing the category {G,m,e,i}?

Perhaps C = Sets??

  — Preceding unsigned comment added by Tekolste (talk • contribs) 21:17, 24 February 2016 (UTC)[reply]

...I will either come to grips with the current wording or offer something better. — Preceding unsigned comment added by Tekolste (talk • contribs) 18:11, 24 February 2016 (UTC)[reply]

It is also quite weird - we're in an arbitrary category, but the example defines "functions (morphisms of set in a category)" and saying that "The identity function e sends each element in * to 1" which is plainly wrong in arbitrary categories - we don't know anything about the structure of *, it doesn't necessarily have to be a set.

141.3.238.8 (talk) 14:12, 20 January 2017 (UTC)[reply]

I am removing the offending sentence now. A copy of it is below.

Let C be an arbitrary category with finite products and a terminal object *. The group G is an object in C. The morphism e sends each element in * to 1, the identity element of the group G.

The above is trying to say something that almost makes sense, but I cannot quite figure out what was intended. Perhaps this?

Let C be a category with finite products and a terminal object *, such that the group G is an object in C.

OK that makes sense. But then the last sentence: * does not have "elements", but perhaps e sends morphisms

C\ni c\to *

to 1? So the domain of e are morphisms? And the codomain is G? Ugh. What's the point of this? The last version which made sense to me was the version of 26 October 2014; I am restoring that version now. 67.198.37.16 (talk) 03:59, 18 March 2021 (UTC)[reply]

I restored the older version and made some minor cleanup edits; I think this resolves all of the confusion immediately above. 67.198.37.16 (talk) 04:43, 18 March 2021 (UTC)[reply]

Most of the examples look to be incorrect[edit]

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)[reply]

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)[reply]

Removed laws from the Groups section.[edit]

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 $\forall xy,m(x,i(y))=e(*)$ , 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)[reply]