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Given $U,V \in Vect$ we can build the tensor product as $U \otimes V := F(U \times V) / F(N)$ where $F:Set \rightarrow Vect$ is the free functor and where

$N:={(au,v) - a(u,v), (u,av) - a(u,v), (u+u',w) - (u,w) - (u',w), (u,v+v') - (u,v) - (u,v') : a \in K; u, u' \in U; v, v' \in V }$

Then we have the projection $p:U \times V \rightarrow U \otimes V$. The universalism of the projection, says that it is initial amongst all morphisms whose kernel includes $F(N)$. This means that these morphisms will also be bilinear. Thus we see that the tensor morphism is initial amongst all morphisms that are bilinear.

How does one phrase this in the terminology of universal morphisms? The Wikipedia page gives an example of the tensor algebra where they start with a forgetful functor $U:K-Alg \rightarrow K-Vect$.

I don't see how something similar can work here since the tensor product is an element of $Vect$. Yet it seems to me that the notion of a universal morphism here is appropriate.

Mozibur Ullah
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    Wikipedia gives the universal property of tensor products (of vector spaces). Is there some issue with the characterization it gives? – Derek Elkins left SE Feb 01 '18 at 00:56
  • @Derek Elkins: This is the same universal property that I referred to in the question ('we see that the tensor morphism is initial'). What I'm asking about is how to fit this in the definition of a universal morphism. – Mozibur Ullah Feb 01 '18 at 01:23
  • It does list the tensor product as an example but it doesn't give details ... – Mozibur Ullah Feb 01 '18 at 01:30
  • Quoting that Wikipedia page: "The term universal morphism refers either to an initial morphism or a terminal morphism, and the term universal property refers either to an initial property or a terminal property." In general, initiality, universal arrows, and also representability are equivalent notions. Most category theory introductions will describe all three. I explicitly focus on these and their relationship in this blog post. You should learn to move between these different presentations of universal properties. – Derek Elkins left SE Feb 01 '18 at 02:01
  • @Derek Elkins: sure - and that's why I'm asking this question. There's no need to be patronising by quoting an online Encyclopedia. If you are comfortable about moving from these different descriptions then perhaps you can describe how it's done for the tensor product - as I can't see it. I've just looked at the description of the categorical product as a terminal morphism that the author has written as an example and some of the details is wrong. Plus it's not a full description. – Mozibur Ullah Feb 01 '18 at 02:34
  • But that's Wikipedia for you ... it has it's flaws as well as it's strengths. – Mozibur Ullah Feb 01 '18 at 02:36
  • That example, by the way, won't work here as the categorical product in vector spaces will be the usual product of vector spaces. Any pointers? – Mozibur Ullah Feb 01 '18 at 02:39
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    The reason I quoted the Wikipedia page is you say (twice now) that "[you] see that the tensor morphism is initial" and the definition of "universal morphism" that you seem to be using is the one from Wikipedia that defines "universal morphism" as an "initial morphism". That makes it confusing for you to ask how to describe something as a universal morphism when you seem to be saying you already see how it is an initial morphism. Perhaps you could spell out exactly what "the tensor morphism is initial" means to you. Then attempt in detail to give a universal morphism and say where you're stuck. – Derek Elkins left SE Feb 01 '18 at 04:06
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    There are several areas where you seem confused. Maybe you're not, but it is hard to tell. First, a universal property doesn't provide a construction, only a characterization. Second, the tensor product is not a tensor algebra nor a categorical product, so neither of those examples on the Universal Property Wikipedia page are relevant. Third, none of the details are wrong in the description of the categorical product on the Wikipedia page nor do I know what you think is missing such that "it's not a full description" unless you're (incorrectly) expecting a construction. – Derek Elkins left SE Feb 01 '18 at 04:06
  • @Derek Elkins: Au contraire. First of all, I'm asking a question to sort out the confusions that I'm having specifically about the tensor product. I'm not confused about the other things that you mention. Perhaps, you'll notice that although a universal property of the tensor product doesn't provide a construction I do provide a construction by using the universal property of the quotient of modules and I show that this satisfies the universal property of the tensor product. The example in question on Wikipedia is wrong. Have a look at the last line, they have the morphisms the wrong way. – Mozibur Ullah Feb 01 '18 at 04:36
  • To add, I expected people who knew enough to advise me on this would know that the projection on the quotient is initial, so there was no need to spell that out in detail. – Mozibur Ullah Feb 01 '18 at 04:44
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    I checked again, and the Wikipedia page doesn't have the morphisms the wrong way (though I will admit the last sentence of that section is poorly written). Can you write out exactly what you think is wrong in that section and what you think it should be? – Derek Elkins left SE Feb 01 '18 at 05:07
  • You have some confusion somewhere; I'm trying to pinpoint where that confusion is. If you can spell out the details for why the "tensor morphism is initial", then I and others will know it's not there. "I don't see how something similar can work here since the tensor product is an element of Vect" is the only indication you've (explicitly) given of where you're confused, but it is not at all clear why you think the tensor product being "an element of Vect" is a problem. Can you write out what you have tried to solve your problem? Or, feel free to continue wasting time insulting me. – Derek Elkins left SE Feb 01 '18 at 05:09
  • I'm afraid the confusion is all on your side. The quotient morphism is initial, I constructed the tensor product by the quotient morphism: ergo, it is initial. By the way, the example I mentioned had the right idea, but it was faultily executed. The devils always in the detail...I'm not trying to insult you, I'm just a little irritated by your patronising tone. I think that's fair enough under the circumstances. – Mozibur Ullah Feb 01 '18 at 05:21
  • @Derek Elkins: I said 'I don't see how this could work here' because $U$ is a forgetful functor, so I couldn't see how a functor $Vect \rightarrow Vect$ could be seen as forgetful, there's no structure being forgotten. In the example I gave, the tensor algebra, the functor is forgetful: it forgets the algebra structure. – Mozibur Ullah Feb 01 '18 at 05:28
  • Are you looking for a purely category-theoretic description of the tensor product, only in the language of objects and arrows, etc? Is that what you're asking for? – Mike Feb 01 '18 at 18:49
  • @mike: I've already provided the description of the tensor product in terms of objects and arrows, its a formal description in terms of a universal arrow that I'm looking for. – Mozibur Ullah Feb 03 '18 at 05:32
  • I don't think you have. Without referring to specific categories, tensor product is usually dealt with (axiomatically) by adding additional (monoidal) structure to an abstract category. – Mike Feb 03 '18 at 08:59
  • @mike: Really? I think you're getting things a little mixed up simply because the word tensor appears in both contexts. A tensor or monoidal category is additional structure on an abstract category but it has nothing directly to do with the tensor product. It's an axiomatisation of some properties of the tensor product. Please reread the second paragraph - and tell me why this isn't a characterisation in terms of morphisms. – Mozibur Ullah Feb 03 '18 at 10:04
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    You seem to like telling people that they are confused. I am answering to the title. Monoidal categories is the answer to that. If you are simply looking for a $\textbf{universal arrows}$ description of tensor product, then you should read about the tensor-hom adjunction. As every adjunction's unit is a family of universal arrows. – Mike Feb 03 '18 at 11:55
  • @Mike: Hmmm....I didn't say that you did. The evidence however speaks for itself. – Mozibur Ullah Feb 03 '18 at 12:12
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    I couldn't agree more – Mike Feb 03 '18 at 12:15
  • Monoidal categories aren't the answer. Nor did you answer my question 'as to why this isn't a characterisation in terms of morphisms' given your pointed insistence 'I don't think you have', which to be honest - is just patronising. – Mozibur Ullah Feb 03 '18 at 12:18
  • @Derek Elkins:You might be interested in the solution I give, given your comments. – Mozibur Ullah Feb 06 '18 at 18:31
  • @Mike: You might be interested in my solution to the problem, which as I pointed out to you, which ought not to, and in fact does not rely on the theory of monoidal categories. – Mozibur Ullah Feb 06 '18 at 18:33

1 Answers1

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The solution relies upon the fact that the tensor product is the coproduct in the category of commutative algebras. A proof of this is here in an answer in Math.SE by Omar Antolin-Carmenera.

Then we need only note that the categorical product has a formulation as a universal morphism, in fact, a terminal morphism as pointed out in the second example on the Wikipedia article on universal morphisms (note that there is a mistake in the detail of the proof, though the idea is the right one).

Finally, we just need take the dual of this construction to formulate the coproduct as an initial morphism.

Mozibur Ullah
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