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?

Algebra(Chapter II, No. 4, Proposition 4), where it is proved that if one of the pairs (A,B), or (A,C), or (B,D) consists offiniteprojective modules, then the canonical map in question is in fact bijective. – Mahdi Majidi-Zolbanin Aug 5 '11 at 4:19