Proof of the properties of tensor product

Question

On page 25 of Atiyah-Macdonald "Introduction to commutative algebra", the author says that

"We shall never again need to use the construction of the tensor product given above and the reader may safely forget it if he prefers. What is essential to keep in mind is the defining property of the tensor product."

I am not sure whether I understand that part well. When we prove some properties tensor product, we typically use the generators (pure tensors). It seems to me that that kind of step makes use of the construction of tensor product.

So my question is, is there a purely categorical proof (without mentioning elements, only using universal property) for various properties of tensor product? (Even for the simplest one like $M\otimes N$ $\cong N\otimes M$, all the proofs I have seen involve elements when checking the two maps made from universal property are inverses to each other.)

If it is, is it possible to use these kind of method to the other categories than the category of modules? (For example, for the tensor product of sheaves.)

Answer 1

$\newcommand{\tensor}{\otimes}\newcommand{\from}{\colon}$Let $M \tensor N$ be an $A$ module, and $f \from M \times N \to M \tensor N$ be a bilinear map that satisfy the universal property of the tensor product. The elementary tensor $m \tensor n$ is defined to be $f(m,n)$. We will prove using just the universal property and not the construction that:

$M \tensor N$ is spanned by the elements $m \tensor n$ as $m$ varies over $M$ and $n$ varies over $N$.

Let $P$ be the submodule generated by the elementary tensors. Then by bilinearity, the image of $f$ actually lies in $P$. Let us denote by $f'$ the map $M \times N \to P$, which agrees with $f$ everywhere. Let $i$ denote the inclusion $P \to M \tensor N$.

Now, $f = i \circ f'$ and $f'$ is bilinear. Hence there is a unique map $g \from M \tensor N \to P$ such that $g \circ f = f'$. Now, $g \circ i = 1_P$, since $$g(m \tensor n) = g(f(m,n)) = f'(m,n) = m \tensor n \ .$$ Finally, for $m \in M$, $n \in N$, we have that $$ i(g(m \tensor n)) = i(m \tensor n) = m\tensor n\ ,$$ and hence $i \circ g \circ f = f = 1_{M \tensor N} \circ f$. Now, considering that $f$ being a bilinear map, factors through itself in a unique way, $i \circ g = 1_{M \tensor N}$. In particular, $i$ is surjective, which is to say $P = M \tensor N$.

Answer 2

There is a proof that the universal property determines the tensor product of modules up to isomorphisms. The initial construction is only used to prove the existence of the tensor product. It is not a bad thing to visualize by tensors but sometimes it is better to strictly use the universal property. $\require{AMScd}$ \begin{CD} A\times B @>\tau>> A\otimes B\\ @V f V V\# @V \exists !\varphi VV\\ B\otimes A @= B\otimes A \end{CD} Given a bilinear function $f$, there is a unique morphism $\varphi$ that makes the diagram commute and that must be the morpism that maps $b\otimes a\mapsto a\otimes b$. And by interchange $A\otimes B$ and $B\otimes A$ in the diagram you also get a unique morphism $\varphi'$. And $\varphi\varphi'=1_{B\otimes A}$, $\varphi'\varphi=1_{A\otimes B}$.