Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why?

Question

Sorry for asking a linear algebra question on a research forum, but this seems to be either a case of extreme blindness on my side, or a case of a result lying much deeper than it seems.

The following theorem is "easily seen" according to a text I have been reading (more precisely, it is part of Proposition I.1.2 in that text):

Theorem 1. Let $A$, $B$, $C$, $D$ be four vector spaces over a field $k$. Then, the canonical map

$\mathrm{Hom}\left(A,C\right)\otimes\mathrm{Hom}\left(B,D\right) \to \mathrm{Hom}\left(A\otimes B,C\otimes D\right)$,

$f\otimes g\mapsto \left(a\otimes b\mapsto f\left(a\right)\otimes g\left(b\right)\right)$

is injective.

I see how this is trivial if $A$ and $B$ are finite-dimensional. I also see that it is indeed easy if $C$ and $D$ are finite-dimensional. But without finite-dimensionality conditions I have nowhere to start. The $\mathrm{Hom}$ functor does not commute with direct sums, while $\otimes$ does not commute with direct products (or does it over a field?), so there seems to be no easy way to reduce it to finite-dimensional cases. How can we proceed then?

Also, is there any application of the above theorem outside of the two cases I mentioned?

To make this more interesting, how much is saved if we let $k$ be a commutative ring with $1$, and require (say) flatness instead of freeness?

Answer 1

Suppose $\sum f_i\otimes g_i$ is in the kernel and assume that the $g_i$ are linearly independent. For every $a\in A$ and $\lambda\in C^*$, we have $\sum\lambda(f_i(a))g_i=0$, so by assumption $\lambda(f_i(a))=0$ for all $i$. Since $a$ and $\lambda$ were arbitrary, this implies $f_i=0$ for all $i$.

Answer 2

Here is a alternative proof, which is not as short as a-fortiori's but it seems to be more conceptual and includes the following lemma which is useful in its own right: (Everything takes place over a field)

Lemma. The natural map $V \otimes \prod_i W_i \to \prod_i (V \otimes W_i)$ is injective.

Proof: Choose a basis $B$ of $V$. Then the map corresponds to the natural map $(\prod_i W_i)^{(B)} \to \prod_i (W_i^{(B)})$, given by a kind of transposition $((w_{ib})_{i})_{b} \mapsto ((w_{ib})_{b})_{i}$, obviously injective.

Corollary. The natural map $\prod_i V_i \otimes \prod_j W_j \to \prod_{i,j} V_i \otimes W_j$ is injective.

Proof: Apply the Lemma twice.

Lemma. The natural map $\hom(V',V) \otimes \hom(W',W) \to \hom(V' \otimes W',V \otimes W)$ is injective.

Proof: Choose basis $B,C$ of $V',W'$. Then the map corresponds to the natural map $V^B \otimes W^C \to (V \otimes W)^{B \times C}$, which is injective by the Corollary.

Answer 3

I think this boils down to the fact that the tensor product of two nonzero vector spaces over a field is nonzero. Let $C_1$ be the image of $f$ and $D_1$ be the image of $g$. Then $C_1$ is a sub-vector space of $C$ and $D_1$ is a sub-vector space of $D$. Let's denote the canonical map by $\theta$. If $f\otimes g\in\ker\theta$ we want to show $f\otimes g=0$. By definition of $\theta$, to say $\theta(f\otimes g)=0$ means that for all $a\in A$ and for all $b\in B$, $f(a)\otimes g(b)=0$. This would imply that $C_1\otimes D_1=0$. Thus either $C_1=0$ or $D_1=0$.