Tensor products and natural isomorphisms Let $U$ and $V$ be two finite-dimensional vector spaces.  How to prove that $U \otimes V \cong V \otimes U$?
The equivalence symbol generally refers to natural isomorphisms – i.e. isomorphisms defined without any reference to the representation of the underlying vector spaces. This is the point that I try to understand.
A straightforward proof is derived from the universality property of the tensor product definition. Once the bilinear function $h_1 : U \times V \rightarrow U \otimes V$ (from the tensor definition) and $h_2 : U \times V \rightarrow V \otimes U$ (from the tensor definition) are defined, there is a unique linear function $f$ between $U \otimes V$ and  $V \otimes U$. BUT this function depends on $h_1$ and $h_2$. How can we conclude there is a natural isomorphism between both vector spaces?
A related problem can be found in the proof that two tensor products definitions are isomorphic. Again the proof relies on the universality property of the tensor product definition. Thus, the isomorphism between the two tensor products depends on $h_1$ and $h_2$.
Is there any explanation on the aspect of the natural isomorphism without relying on category theory (and functors and Yoneda lemma) or defining the tensor product  “as the quotient space of a vector space having U × V as a basis”? Many references mention the natural isomorphism, use the universality property, but never mention those more formal set-ups.
 A: Wasn’t category theory developed to make sense of the word natural isomorphism? Anyway this is more of a lengthy comment than an answer, but maybe it helps anyway.
As far as I am concerned there can only be one definition of the tensor product and the other characteristic properties are, well, properties. What is natural/canonical depends on the definition.
If you define the tensor product $U \otimes V$ as a quotient, you have an explicit generating system consisting of pure tensors $u\otimes v$. Without invoking the universal property you may check that the natural/canonical = obvious/immediate assignment
$$\begin{array}{ccc}
U \otimes V & \rightarrow & V \otimes U\\
u \otimes v & \mapsto & v \otimes u
\end{array}$$
extends to a welldefined isomorphism $U \otimes V \cong V 
\otimes V$.
If you define the tensor product via the universal property (note that the bilinear map $U \times V \rightarrow U \otimes V$ is part of the datum of a tensor product!), you are doing category theory. In this case canonical/natural is commonly used to denote morphisms induced by universal properties. But that means in particular that you need to have some data, by which they are induced. In our case this happens by using the bilinear maps $h_1,h_2$ you mention. Note that they can be defined in a canonical way as well in that the universal property of the product gives a canonical isomorphism $U \times V \cong V \times U$ and $h_1,h_2$ are just bilinear maps obtained from postcomposing this isomorphism with the bilinear map defining the tensor product.
The point of the latter perspective is that we don’t need to specify a basis and use linear combinations, extending linear maps etc. In other words it is independent from any particular representation of our vector spaces as $K^I$. It is not about not having to specify anything and getting isomorphisms for free. It is rather about getting things with minimal (but nonzero!) effort...
