Understanding the tensor product of functions/maps Let $K^A = \{f \,|\, f\colon A \longrightarrow K\}$ be the set of functions from $A$ (some arbitrary set) to $K$ (a field, I guess, I want to be able to call $K^A$ a $K$-vector space). Given some other set $A'$ I'd like to know when can we say
$$K^A\otimes K^{A'} \subseteq K^{A\times A'}$$
and when can we say they're equal, or when is this true for subalgebras (like $C^\infty(A)$, $L^2_\mu(A)$ or $A^* = \mathrm{Hom}(A;K)$, for instance, depending on the structure $A$ has). The tensor product of functions would be define so that
$$(f\otimes g)(x,y) = f(x)g(y).$$
I tried doing some research and piecing together what I've learnt so far. Most resources don't talk about purely algebraic vector spaces, though, and I don't know much about Banach spaces, functional analysis or real-variable analysis and it's being a little hard.

*

*For the case of linear forms (so $A,A'$ are $K$-vector spaces) it is true that their tensor product yields a "bilinear form" $A\times A'\longrightarrow K$, so $A^*\otimes A'^* \equiv \mathrm{Bil}(A,A';K)$.


*It can't be true in general, because rank-1 elements of the tensor product correspond to (product-wise) separable functions and not every function is a linear combination of separable functions. For example, the function $\delta\colon \mathbb{R}\times\mathbb{R} \longrightarrow \mathbb{R}$ defined to be $0$ everywhere except $1$ wherever $x = y$ is not a linear combination of separable functions.


*For $L^2(A)$, at least when $A\subseteq \mathbb{R}^n,A'\subseteq\mathbb{R}^m$ with the Lebesgue measure, it isn't true that
$$L^2(A)\otimes L^2(A') = L^2(A\times A')$$
but it is true if we use the "Hilbert" tensor product (the closure of the usual tensor product with respect to the topology induced by the metric). This is because the topology of the Hilbert space allows for "infinite linear combinations" to make sense whenever they're convergent.
I know this question might be too broad (sorry!), but I'm sure there's some concept to know or reference to read from which I could learn more about this.
 A: I gave a long detailed answer to a related question on MO at
https://mathoverflow.net/questions/363935/what-is-the-role-of-topology-on-infinite-dimensional-exterior-algebras/364211#364211
but, since it seems that on math.stackexchange the preference is to have self-contained answers, let me say again here part of what I explained on MO.
The first task is to understand the algebraic tensor product $K^A\otimes K^{A'}$. The general construction proceeds via the free vector space with basis indexed by symbols $f\otimes g$ with $f\in K^{A}$, $g\in K^{A'}$ and quotienting by relations $(f_1+f_2)\otimes g-f_1\otimes g-f_2\otimes g$ etc. Another equally uninspiring construction is to take an (uncountable if $A$ is infinite, even countably so) Hamel basis $(e_i)_{i\in I}$, for $K^{A}$ and similarly a Hamel basis $(f_j)_{j\in J}$ for $K^{A'}$, produced by the Axiom of Choice, and realize $K^A\otimes K^{A'}$ as the subset of $K^{I\times J}$ made of functions of finite support (equal to zero except for finitely many elements of $I\times J$). However, the proper definition is as the solution to a universal problem: $K^A\otimes K^{A'}$ together with a bilinear map $\otimes:K^{A}\times K^{A'}\rightarrow K^{A}\otimes K^{A'}$ must be such that for every vector space $V$ and bilinear map $B:K^A\times K^{A'}\rightarrow V$, there should exist a unique linear map $\varphi:K^A\otimes K^{A'}\rightarrow V$ such that $B=\varphi\circ\otimes$.
The main point I want to make in relation with the OP's question is: one can construct such a space concretely as follows.
Let $W$ be the subset of $K^{A\times A'}$ made of functions $h:(x,y)\mapsto h(x,y)$ which are finite sums of functions of the form $f\otimes g$ with $f\in K^{A}$, $g\in K^{A'}$. Here $f\otimes g$ is the function $A\times A'\rightarrow K$ defined by
$$
(f\otimes g)(x,y)=f(x)g(y)
$$
for all $x\in A, y\in A'$. Note that the definition also provides us with a bilinear map $\otimes:K^A\times K^{A'}\rightarrow W$.
Proposition 1: The algebraic tensor product $K^{A}\otimes K^{A'}$ can be identified with $W$.
In other words, $W,\otimes$ solves the mentioned universal problem.
The proof relies on the following lemmas.
Lemma 1: For $p,q\ge 1$, suppose $e_1,\ldots,e_p$ are linearly independent elements in $K^{A}$ and suppose $f_1,\ldots,f_q$ are also linearly independent elements in $K^{A'}$. Then the $pq$ elements $e_a\otimes f_b$ are linearly independent in $W$.
Proof: Suppose $\sum_{a,b}\lambda_{a,b}e_a\otimes f_b=0$ in $W$. Then $\forall (x,y)\in A\times A'$,
$$
\sum_{a,b}\lambda_{a,b}e_a(x) f_b(y)=0\ .
$$
If one fixes $y$, then one has an equality about functions of $x$ holding identically.
The linear independence of the $e$'s implies that for all $a$,
$$
\sum_{b}\lambda_{a,b}f_b(y)=0\ .
$$
Since this holds for all $y$, and since the $f$'s are linearly independent, we get $\lambda_{a,b}=0$ for all $b$. But $a$ was arbitrary too, so $\forall a,b$, $\lambda_{a,b}=0$ and we are done.
Lemma 2: Let $B$ be a bilinear map from $K^A\times K^{A'}$ into some vector space $V$.
Suppose $(g_k,h_k)$, $1\le k\le n$ are elements of $K^{A}\times K^{A'}$ satisfying
$$
\sum_{k}g_k\otimes h_k=0 
$$
in $W$, i.e., as functions on $A\times A'$. Then
$$
\sum_k B(g_k,h_k)=0
$$
in $V$.
Proof: This is trivial if all the $g$'s are zero or if all the $h$'s are zero. So pick a basis $e_1,\ldots,e_p$ of the linear span of the $g$'s and pick a basis $f_1,\ldots,f_q$ of the linear span of the $h$'s (no Axiom of Choice needed). We then have decompositions of the form
$$
g_k=\sum_a \alpha_{k,a}e_a
$$
and
$$
h_k=\sum_b \beta_{k,b} f_b
$$
for suitable scalars $\alpha$, $\beta$.
By hypothesis
$$
\sum_{k,a,b}\alpha_{k,a}\beta_{k,b}\ e_a\otimes f_b=0
$$
and so $\sum_k \alpha_{k,a}\beta_{k,b}=0$ for all $a,b$, by Lemma 1.
Hence
$$
\sum_k B(g_k,h_k)=\sum_{a,b}\left(\sum_k \alpha_{k,a}\beta_{k,b}\right) B(e_a,f_b)=0\ .
$$
Now the proof of Proposition 1 is easy. The construction of the linear map $\varphi$ proceeds as follows.
For $v=\sum_{k}g_k\otimes h_k$ in $W$, we let $\varphi(v)=\sum_k B(g_k,h_k)$.
This is a consistent definition because if $v$ admits another representation $v=\sum_{\ell}r_{\ell}\otimes s_{\ell}$, then
$$
\sum_k g_k\otimes h_k\ +\ \sum_{\ell}(-r_{\ell})\otimes s_{\ell}=0
$$
and Lemma 2 implies
$$
\sum_k B(g_k,h_k)=\sum_{\ell} B(r_{\ell},s_{\ell})\ .
$$
The other verifications that $W$ with $\otimes$ solve the universal problem for the algebraic tensor product pose no problem. So, I think this answers in the affirmative the question about the inclusion of $K^{A}\otimes K^{A'}$ inside $K^{A\times A'}$. It is not hard to see that the inclusion is strict unless $A$ or $A'$ is finite.
Again more details, especially when also taking completions and bringing topology into the game, can be found in the linked MO answer. In particular, I worked out the identification of the projective tensor product $\ell^1(\mathbb{N})\widehat{\otimes}_{\pi}\ell^1(\mathbb{N})$ with $\ell^1(\mathbb{N}\times\mathbb{N})$. The same proof also works for $\ell^p$ spaces with $p\in [1,\infty)$.
Finally, when considering some kind of function spaces $\mathscr{F}(A)$ on geometric spaces $A$ say like $\mathbb{R}^n$ or nice manifolds, one can ask if within this category one has
$$
\mathscr{F}(A)\widehat{\otimes}\mathscr{F}(A')\simeq \mathscr{F}(A\times A')
$$
for a suitable notion of topological tensor product.
While Banach spaces are more accessible to beginners and may seem like a good choice to see this kind of identities in action, the best setting is rather that of spaces of Schwartz distributions and test functions. This relates to the comment by Paul (see his vignette I linked to in my MO answer for more details).
