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What is this?
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 . The object U(S) is "universal" for S, in the sense that, for any H in G and any map , there exists a unique homomorphism such that .
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)
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
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 : X → Y. The problem is there doesn't seem to be a sensible choice for l : K → Y. 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)
I don't understand this particular example of universal morphism. It says that is a universal morphism for with respect to the diagonal map . But is not defined on , as it should be, according to the definition of universal morphism.—Preceding unsigned comment added by 18.104.22.168 (talk) 14:03, 18 September 2008 (UTC)
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)
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.  , 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 (really, it's the comma category , 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 22.214.171.124 (talk) 03:49, 28 September 2011 (UTC)
φ vs. Φ
- That's not an upper-case Phi, but a different variant of lower-case phi. Should be corrected anyway. -- 126.96.36.199 (talk) —Preceding undated comment added 11:50, 17 August 2012 (UTC)
Object "A" in initial morphism really unique up to isomorphism?
In the last paragraph under the subsection "Existence and uniqueness" is the statement "The object A itself is only unique up to isomorphism." It makes sense that in the category of morphisms from X to U that the pairs (A,phi) and (A',phi') are isomorphic, as sketched in the previous paragraph, but I believe that A and A' can be nonisomorphic in their category D.
For example take C the category of three objects X, A, A', with nonidentity morphisms being phi:X->A, phi':X->A', g_12:A->A', g_21:A'->A, h_1:A->A, and h_2:A'->A' (and all powers of h_1,h_2). Make it so that (g_12 o g_21) = h_2 and (g_21 o g_12) = h_1. Then take D the subcategory consisting of A and A', and let U be the functor taking objects and morphisms to themselves. We see that A and A' are not isomorphic: the only maps between them compose to something not the identity. And we can make sure that (A,phi) and (A',phi') are initial morphisms if we make sure (h_1 o phi) = phi and (h_2 o phi') = phi'.
Is something wrong in my counterexample above or is the statement in the article incorrect?
Morphisms from object to functor
In the "Formal definition" section, we speak of the "category of morphisms from to " ( and categories, an object of , and a functor from to ), with a wikilink to the "Comma category" page. But on the "Comma category" page there is no mention of such a thing. There we define for functors and ; the slice and coslice categories and for an object of category ; "the discrete category whose objects are morphisms from to " for objects and of a category.)
Is this thing a standard notation? What is it? (My best guess is that it's where is a morphism from the category with one object, , to with .) Can we explain it properly within this article ("Universal property")? Buster79 (talk) 15:00, 27 November 2015 (UTC)
From Cgibbard's comment above, "(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)", so my guess was OK. I think it would be nice for this page to be more self-contained. Can't see how anyone who doesn't already know a lot about category theory could make much sense of it. If we give a more elementary definition of "initial morphism" and //then// remark that it could also be defined as an initial object in a certain comma category, things would be a lot clearer. Buster79 (talk) 15:09, 27 November 2015 (UTC)