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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?

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Regarding the second part of your question, where $k$ is a commutative ring with $1$ and $A,B,C,D$ are $k$-modules, you can find a partial answer in Bourbaki's 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 of finite projective modules, then the canonical map in question is in fact bijective. –  Mahdi Majidi-Zolbanin Aug 5 '11 at 4:19
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Thanks. I assume this easily follows from "finite projective modules are direct addends of finite free modules"., –  darij grinberg Aug 5 '11 at 8:57

3 Answers 3

up vote 26 down vote accepted

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 $i$.

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Wonderful proof! –  darij grinberg Aug 3 '11 at 18:18
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I think this is the only trick I know for proving that something in a tensor product is zero. Fortunately, it always works. –  Anton Geraschenko Aug 3 '11 at 18:21
    
There is a general criterion when an element in a tensor product of modules $M \otimes N$ is zero: Let $E$ be a generating set of $M$ and $F$ be a generating set of $N$, then a typical element of $M \otimes N$ has the form $\sum_{e \in E} e \otimes n_e$. It vanishes if and only if there is a matrix $(\lambda_{e,f})_{e,f}$ over $R$, whose support is finite, such that $n_e = \sum_{f \in F} \lambda_{e,f} f$ for all $e$ and $0 = \sum_{e \in E} \lambda_{e,f} e$ for all $f$. –  Martin Brandenburg Aug 5 '11 at 18:58

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.

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By the way, this also shows directly that the maps in consideration are not surjective in the infinite-dimensional case; it even gives a "column-finite" type description when we choose bases. So let me add this side question: Is there an intrinsical description of the image of $\hom(V',V) \otimes \hom(W',W)$ in $\hom(V' \otimes W',V \otimes W)$? For example the image of $V \otimes W^* \to \hom(W,V)$ consists of the homomorphisms of finite rank. –  Martin Brandenburg Aug 5 '11 at 9:11
    
Nice proof, Martin. (I don't have the time, right now, to search for the intrinsical description though.) –  darij grinberg Aug 5 '11 at 11:07

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 $D_1=0$.

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My answer is not quiet correct, I should have started with a linear sum of $f_i\otimes g_i$'s. –  Mahdi Majidi-Zolbanin Aug 3 '11 at 18:15

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