Talk:Universal property

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Mathematics (Rated Start-class, High-importance)
WikiProject Mathematics
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
Start Class
High Importance
 Field: Foundations, logic, and set theory

What is this?[edit]

I was reading something (that didn't use categorical language) that gave an informal definition of what clearly must be a universal property, and I thought, just for grins, I'd match it up to the definition given in this article. Much to my horror, I couldn't make it work. What am I doing wrong?

The claim went like this: Given an object S in category M, there exists a corresponding object in category G, called U(S). Structure is preserved, in that there is a homomorphism \mu:S\to U(S). The object U(S) is "universal" for S, in the sense that, for any H in G and any map \phi:S\to H, there exists a unique homomorphism \psi:U(S)\to H such that \psi \circ \mu = \phi.

No matter how I tried, I could not make this fit into the "universal" or "co-universal" definitions, yet clearly its universal.

On closer examination, the above is a lot like the definition of a tensor algebra, in that I can relabel the letters (S is V, U(S) is TV, G is the category of algebras, etc.). However, I cannot relabel the letters so as to fit the diagrams given in the formal definition in the first few paragraphs. Similarly, I can't make the tensor-algebra definition fit either. linas 15:57, 24 March 2007 (UTC)

Well, I note that the tensor algebra description has become rather muddled. It is clearer is some older revisions. Basically, if V is a vector space and U : VectK → AlgK is the forgetful functor then the pair (TV, i : V → TV) is a universal morphism from V to U.
I can't make any sense of your problem however. You have "morphisms" between objects in different categories, which is, of course, nonsense. Are you implicitly using some forgetful functor or other identification? -- Fropuff 02:17, 25 March 2007 (UTC)
Sorry, I meant "morphism" in the ordinary sense (homomorphism), not the category sense. I understand the tensor algebra bit perfectly well; that is not the problem. So, if you take what I wrote, and replace the letter S with the letter V, and replace the string U(S) with TV, etc. you will get exactly the definition of the tensor algebra. So there's not a problem there; my thing is clearly universal in the exactly the same way that the tensor algebra is. The problem is that I cannot find a way of taking this definition, and substituting letters and strings to get the definition in the article, which uses the letters D,C,X, A, phi, etc. Just try it. Plug in U : VectK → AlgK for U:D→ C, in the article, and then try to fill in the rest of the diagram, and I believe you will see the trouble. linas 04:38, 25 March 2007 (UTC)
The forgetful functor U goes from AlgK to VectK not the other way around. Explicitly, the substitution is C = Vect, D = Alg, U = U, X = V, A = TV, φ = (i : V → U(TV), the inclusion map). It all works perfectly. I think maybe you are confusing the forgetful functor U with the tensor algebra functor T. The only functor that comes into play in the definition is U. The functor T exists by virtue of the fact that the universal object TV exists for all V. -- Fropuff 06:28, 25 March 2007 (UTC)
Yes, that's exactly it, I was confusing T with U, and then wondering why all the arrows seemed to be pointing in all the wrong directions. Thanks, I got it now, my faith is restored. linas 16:16, 25 March 2007 (UTC)

Monads[edit]

Well, my confusion prompted some further reading. If I understand things correctly, then the left adjoints always(?) seem to come in pairs, so that e.g. the free functor is the left adjoint to the forgetful functor. Thus, one (always?!) has a monad (category theory). In the monad, one writes T=FU, builds a T-algebra, and the Eilenberg-Moore category of T-algebras. From what I can tell, the Eilenberg Moore category is always isomorphic to whatever category we started with (provided we started with a "finitary" category or variety of algebras). Right? I'm thinking that this article should at least sketch out the above, in maybe 6 sentences, appropriately qualifying the question marks along the way. The point being that this is essentially where the whole idea of "universal property" heads off to anyway; right now, the article doesn't even mention monad, and somehow seems to get bogged down in the middle. linas 21:50, 7 April 2007 (UTC)

Kernels: flawed example[edit]

The example on kernels has been in this article for a long time, but I've never looked at it too closely. Having now done so, the example appears to be fundamentally flawed. The way the category C is constructed (as the category of morphisms in D) requires a morphism (k, l) from 0KK to f : XY. The problem is there doesn't seem to be a sensible choice for l : KY. Taking it to be 0KY, which seems the only logical choice, doesn't work — the universal property will not be satisfied.

The proper thing to do is to replace the category C with the functor category DJ where J is a category with two objects and two parallel morphisms. However, this makes for a wordy and overly complicated example. I think I will scrap the whole thing and replace it with something easier like products. -- Fropuff (talk) 23:13, 7 January 2008 (UTC)


Cartesian Product[edit]

I don't understand this particular example of universal morphism. It says that (X \times Y, \pi_1 \times \pi_2) is a universal morphism for (X,Y) with respect to the diagonal map \Delta:D \rightarrow D\times D. But (\pi_1,\pi_2) is not defined on \Delta(X \times Y), as it should be, according to the definition of universal morphism.—Preceding unsigned comment added by 189.61.207.73 (talk) 14:03, 18 September 2008 (UTC)

Simpler examples?[edit]

Would the following be an example of a universal property? The (topological) closure of a set S is defined as the intersection of all closed sets that contain S. That is, we have on the one hand the intuitive "construction" of "take S and add all its limit points", and on the other hand the definition "T is the closure of S if for every closed set U containing S, T is a subset of U". Does this count? Would this be an useful example to add to the article? Shreevatsa (talk) 17:33, 9 March 2009 (UTC)

Move suggested articles to the bottom[edit]

What do you guys think?Negi(afk) (talk) 07:42, 19 May 2009 (UTC)

Initial/terminal morphism[edit]

The terms were introduced into the article with this edit in October. Is this standard terminology?

My problem is that Ádámek, Herrlich, Strecker (The Joy of Cats) has a conflicting definition that also appears elsewhere, e.g. [1] [2], but I don't remember seeing the one from this article anywhere. --Hans Adler (talk) 21:26, 25 May 2009 (UTC)

The one on this page is just the standard definition of an initial object, specialised to this particular category of morphisms X \to U (really, it's the comma category X\downarrow U, where X is identified with the constant functor that sends everything to X and all maps to the identity.). I've seen this definition used elsewhere. I have no idea at the moment if the definition in Logics for Coalgebras is in any way equivalent to some special case of this one, or even has anything to do with the usual definition of 'initial' in category theory, but such a connection may well exist. I certainly wouldn't assume that they should have anything to do with one another though. Cgibbard (talk) 16:03, 7 September 2009 (UTC)

In that case it would be nice to add a footnote that mentions the existence of the other definition. Unfortunately I don't really feel qualified to find an appropriate formulation. Hans Adler 08:10, 8 September 2009 (UTC)

I have a Ph.D. in Math and I'm trying to learn category theory. (Thank you for the articles). The initial morphism is defined in terms of a pair (A, phi) where phi: X->U(A) I would find it much more understandable if the morphism was called phi-sub-A: X->U(A) because then it would be clear that the morphism depends on A. Is that correct? Otherwise, it's a mystery to me. Andrius Kulikauskas — Preceding unsigned comment added by 108.200.245.131 (talk) 03:49, 28 September 2011 (UTC)

φ vs. Φ[edit]

Why there are different notions (phi vs. Phi) in text and graphics? 92.224.193.202 (talk) 19:01, 8 September 2009 (UTC)

That's not an upper-case Phi, but a different variant of lower-case phi. Should be corrected anyway. -- 132.231.199.50 (talk) —Preceding undated comment added 11:50, 17 August 2012 (UTC)