Talk:Initial algebra

Technical tag

The intro is WAY TOO HARD to understand. Can someone correct that? --M1ss1ontomars2k4 23:20, 13 May 2006 (UTC)[reply]

Not only is it unnecessarily hard; it seems to me that those readers who understand this page already know what an initial algebra is. So, maybe we can agree that the page should be understandable by people who do not yet know what an initial algebra is. It would help if the article were written by someone who knows no category theory.

Why? Because initial algebras arise naturally and can be defined within universal algebra, which is at a lower level of abstraction than category theory. At the end of such an article on initial algebras one could mention that the algebras with a given signature form a category with these algebras as objects and that the uniqueness up to isomorphism of the initial algebra generalizes to the initial object and that many categories that do not consist of algebras with the same signature also have an initial object. In this way the articles would be useful to readers who do not know category theory. Those who do are likely to know anyway what initial algebras are.

Maarten van Emden (talk) 02:08, 5 February 2010 (UTC)[reply]

endofunctor sending X to 1+X - huh?

What is 1 + $\{\aleph _{\omega },42\}$ ?

Someone was having trouble with this on Talk:F-algebra too. I haven't studied category theory properly so don't take my word for it, but I'm guessing that + is the disjoint union. So to answer your question:

1+\{\aleph _{\omega },42\}=\{(1,0),(\aleph _{\omega },1),(42,1)\}

... or something like that. GilesEdkins 00:30, 21 January 2007 (UTC)[reply]

Yes, it definitely is the disjoint union. 1 is the terminal object in the category Set, of sets and functions, and amounts to any singleton set. As any two singleton sets are isomorphic, we say "the" terminal object. (Equality is too strong a property to ask for almost always.) Any morphism 1 ---> X is just an element of X. The coproduct + has the property that to get a morphism X + Y ----> Z one need only specify morphisms f:X ---> Z and g:Y ---> Z. The resulting morphism X + Y ---> Z is usually denoted [f,g]. It has some properties that you can probably find at the coproduct page.

Way too hard indeed

Help, please, let there be a rewrite! I know a little category theory, which I was hoping I might be able to build up to by just starting with initial algebras. Let's have an article for beginners! (Or at least a section for beginners.) Norman Ramsey (talk) 14:21, 5 May 2014 (UTC)[reply]

This mathematical notation needs explaination: 1 + (-)

what is that? what is (-)? is it a place holder? is it a function? This syntax is not found in the functors article.

My recent updates

I've recently updated this article, but, judging by the discussion above, I see it should be updated more. Since it's not year 2005 now, and almost everybody is familiar with categories, I'm curious if $1+(\_)$ is still a problem for many. Etc. I can expand ad libitum, but is in still necessary?

Vlad Patryshev (talk) 16:40, 2 June 2020 (UTC)[reply]

Vlad said: "almost everybody is familiar with categories". That may be true in your social circle, but it's not true for Wikipedia readers in general. I'm familiar with programming languages, and I came here because initial algebras are relevant to the theory of abstract data types. I have no problem with abstract algebra, but category theory is opaque to me, and many people in my social circle have the same problem. Maarten van Emden says that this topic can be understood at the level of abstract algebra, so that is the kind of treatment I need in order to have any hope of understanding the topic. — Preceding unsigned comment added by 209.183.136.7 (talk) 14:34, 19 October 2020 (UTC)[reply]

"strong" functions?

This term (used here with respect to the Ackermann function) is not defined anywhere, and unfamiliar to me. Can anybody explain it? (Otherwise, I think it should be removed). AmirOnWiki (talk) 18:27, 30 January 2024 (UTC)[reply]