I was looking for the infinite tensor product of vector spaces and in literature (for instance, Atiyah and Macdonald's book), I found it for algebras as the direct limit of finite families of algebras. I was particularly interested in vector spaces and found these lecture notes very helpful. The only problem is, I'm unable to understand the proof of the universal mapping, and therefore I've posted this question here, if someone can help me:
Let $I$ be an infinite set and $\{E_i|i\in I\}$ be a family of vector spaces. For each $i\in I$, fix a non-zero $b_i \in E_i$. Define the following product:
$\prod^t_{i\in I}E_i=\{(x_i)\in \prod_{i∈I}E_i|x_i = t_i$
for all but finitely many i}.Next, for each finite subset $J \subset I$, define the tensor product $E_J =
\bigotimes_{i∈J}E_i$ and for finite $J$ and $K$ such $J \subset K$, define the
maps $\varphi _{KJ} : E_J \to E_K$ as: $$\varphi _{KJ} ((x_i)_{i \in
J}) =(\bigotimes_{i∈J}x_i)\otimes (\bigotimes_{i \in K-J} t_i).$$Then, $(E_J, \varphi _{KJ})_{J \subset K \subset I} $ is a directed
system of vector spaces and let $(\bigotimes^t _{i\in I}E_i, \varphi _{J})_{J\subset I}$ be the direct limit of this system. Then $\bigotimes^t _{i\in
I}E_i$ is defined as the infinite tensor product of the family
$\prod^t _{i\in I}E_i$ (see page 4).
If $\bigotimes^t _{i\in I}E_i$ is the tensor product of the family $\prod^t _{i\in I}E_i$, then it should satisfy the following universal property:
For every vector space $F$ and multilinear mapping $u:\prod^t _{i\in I}E_i \to F$, there exists a unique linear mapping $V: \bigotimes^t _{i\in I}E_i \to F$ such that $V \circ \otimes=U$, where $\bigotimes:\prod^t _{i\in I}E_i \to \bigotimes^t _{i\in I}E_i$ is the tensor map.
In page 5 they proved the above universal property but I'm unable to understand. Can anybody make me understand the universal property?
Best Answer
Suppose you have a multilinear map $u$. Then for any finite $J\subset I$, you have the inclusion $\iota_J: \prod_{i\in J}E_i\to \prod_{i\in I}^tE_i$ (which sends $(x_i)_{i\in J}$ to $(y_i)$ with $y_i = t_i$ if $i\notin J$, $x_i$ else), and clearly $u\circ \iota_J$ is multilinear.
Therefore it factors as $\prod_{i\in J}E_i\to \bigotimes_{i\in J}E_i \overset{v_J}\to F$, by definition of the tensor product for a finite family of vector spaces.
Now we need to check that $v_J$ and $v_K$ are compatible if $J\subset K$. This is quite simple, indeed you have the following commutative diagram :
$\require{AMScd}\begin{CD} \prod_{i\in J}E_i @>>>\bigotimes_{i\in J}E_i \\ @VVV @VVV \\ \prod_{i\in K}E_i @>>> \bigotimes_{i\in K}E_i @>>> F\end{CD}$
where the leftmost vertical arrows adds $t_l$'s for $l\notin J$, the other vertical arrow is $\varphi_{KJ}$. Moreover, the map $\prod_{i\in J}E_i\to \prod_{i\in K}E_i\to F$ is (it's easy to check) $u\circ \iota_J$.
It follows (by unicity in the universal property of the tensor product) that $\bigotimes_{i\in J}E_i\to \bigotimes_{i\in K}E_i\to F$ is $v_J$, but it's also $v_K\circ \varphi_{KJ}$.
So the $(v_J)$'s are compatible and therefore assemble into a map $\bigotimes_{i\in I}^tE_i \to F$
We now only have to check that $\prod_{i\in I}^t E_i\to \bigotimes^t_{i\in I}E_i \to F$ is $u$. But we can check that on each element of the restricted product, and each such element is of the form $y=\iota_J(x)$ for some $x$. Now if $y$ is of this form we have the following commutative diagram
$\require{AMScd} \begin{CD} \prod_{i\in J}E_i @>>>\bigotimes_{i\in J}E_i@>>> F \\ @VVV @VVV @V{id_F}VV \\ \prod_{i\in I}^tE_i @>>> \bigotimes_{i\in I}^tE_i@>>> F\end{CD}$
the rightmost square commutes by definition of the map $\bigotimes_{i\in I}^tE_i\to F$
Our goal is to show that the bottom arrows, applied to $y$, yield $u(y)$. But now this is the same as doing down then right-right on $x$. But by commutativity this amounts to doing right-right then down on $x$. But by definition this sends $x$ to $u\circ i_J(x) = u(y)$. So we are done.
(the uniqueness should be clear)