Talk:Universal property

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[edit] 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 \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)

[edit] Monads

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)

[edit] 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)