Induced linear map on $V_1 \otimes \cdots \otimes V_n$ is an epimorphism only if at least $n - 1$ of the $n$ sets $S_j$ are finite

Question

My book reads

$\textbf{1.1.3.}$ As an $\textit{illustrative example}$ [of the tensor product] we consider the special case where each $V_j$ is the set of all real valued functions on some set $S_j$, $W$ is the set of all real valued functions on the cartesian product $S_1 \times \cdots \times S_n$, and $f(v_1, \dots, v_n)(s_1, \dots, s_n) = v_1(s_1) \cdot v_2(s_2) \dots v_n(s_n)$ whenever $v_j \in V_j$ and $s_j \in S_j$ for $j = 1, \dots, n$. Here the corresponding linear map $g$ [induced by the tensor product] is a monomorphism, as seen by induction with respect to $n$; in case $n = 2$ one readily verifies that $\sum_{k=1}^{m} v_{1,k} \otimes v_{2,k} \notin \ker g$ whenever $v_{j,1}, v_{j,2}, \dots, v_{j,m} $ are linearly independent elements of $V_j$, for $j = 1, 2$. We also observe that $g$ is an epimorphism if and only if at least $n - 1$ of the $n$ sets $S_j$ are finite.

All is clear to me except for the last line. Indeed, if at least $n - 1$ of the $n$ sets are finite, then each $\tau \in W$ has finitely many choices for at least $n - 1$ of each entry of its argument, so that (supposing $S_j$ is finite for all $j \neq k$) $\tau$ can be written as a sum over all these choices, that is, $\tau = g(\sum \chi_{s_1} \otimes \cdots \otimes \chi_{s_{k-1}} \otimes \tau(s_1, \cdots, s_{k-1}, \cdot, s_{k+1}, \cdots, s_n) \otimes \chi_{s_{k+1}} \otimes \cdots \otimes \chi_{s_n})$, hence $g$ is an epimorphism.

I do not know how to show the converse, and would appreciate any help.

Answer 1

Let me write this a bit differently. Let $K$ be a field. For sets $S_1,\dotsc,S_n$ we have a canonical linear map $$g : K^{S_1} \otimes \cdots \otimes K^{S_n} \to K^{S_1 \times \cdots \times S_n},\quad f_1 \otimes \cdots \otimes f_n \mapsto ((s_1,\dotsc,s_n) \mapsto f_1(s_1) \cdots f_n(s_n)).$$ It is always a monomorphism, see here. Direct sums commute with the tensor product, but direct products not necessarily. At least we know that finite direct products coincide with finite direct sums. So if, say, $S_2,\dotsc,S_n$ are finite, we get $$K^{S_1} \otimes K^{S_2} \otimes \cdots \otimes K^{S_n} \cong ((K^{S_1})^{S_2})^{S_2 \cdots} \cong K^{S_1 \times \cdots \times S_n}.$$ And it is checked that the composite is $g$, so it is an isomorphism.

Clearly it is also an isomorphism when one of the sets is empty, then on both sides we have the zero vector space.

Now assume conversely that at least two of the sets $S_i$ are infinite, and also (this assumption is missing in your post) that all $S_i$ are non-empty. We claim that $g$ cannot be an isomorphism. We might as well assume that $S_1$ and $S_2$ are infinite (since the whole situation is symmetric in the indices).

Let us consider $n = 2$ first. Then the canonical linear map $K^{S_1} \otimes K^{S_2} \to K^{S_1 \times S_2}$ is not surjective, as has been explained here. Namely, choose two injections $\mathbb{N} \hookrightarrow S_1$, $\mathbb{N} \hookrightarrow S_2$, written as inclusions, and consider the element $h \in K^{S_1 \times S_2}$ that maps $(p,q) \mapsto \delta_{p,q}$ for $p,q \in \mathbb{N}$, and everything else to zero. Then $h$ doesn't lie in the image of $g$.

The general case should be similar. Choose injections from $\mathbb{N}$ as above, and consider the element of $h' \in K^{S_1 \times \cdots \times S_n}$ that maps $(p,q,s_3,\dotsc,s_n) \mapsto \delta_{p,q}$ and everything else to zero. Again you can check that this doesn't lie in the image. In fact, it can be reduced to the case $n=2$. Assume that $h'(s_1,\dotsc,s_n)=\sum_i \lambda_i f_1^i(s_1) \cdots f_n^i(s_n)$. For some fixed elements $s_3,\dotsc,s_n$ (here we use that the sets are non-empty) we then find a corresponding equation for $h(s_1,s_2)$.