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.

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    $\begingroup$ Wikipedia gives the universal property of tensor products (of vector spaces). Is there some issue with the characterization it gives? $\endgroup$ – Derek Elkins Feb 1 '18 at 0:56
  • $\begingroup$ @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. $\endgroup$ – Mozibur Ullah Feb 1 '18 at 1:23
  • $\begingroup$ It does list the tensor product as an example but it doesn't give details ... $\endgroup$ – Mozibur Ullah Feb 1 '18 at 1:30
  • $\begingroup$ 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. $\endgroup$ – Derek Elkins Feb 1 '18 at 2:01
  • $\begingroup$ @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. $\endgroup$ – Mozibur Ullah Feb 1 '18 at 2:34

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.


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