Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:28:37Z http://mathoverflow.net/feeds/question/72013 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72013/homa-c-homb-d-injects-into-homab-cd-when-why Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? darij grinberg 2011-08-03T17:17:31Z 2011-08-05T09:04:31Z <p>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.</p> <p>The following theorem is "easily seen" according to <a href="http://arxiv.org/abs/math/0408405" rel="nofollow">a text I have been reading</a> (more precisely, it is part of Proposition I.1.2 in that text):</p> <p><strong>Theorem 1.</strong> Let $A$, $B$, $C$, $D$ be four vector spaces over a field $k$. Then, the canonical map</p> <p>$\mathrm{Hom}\left(A,C\right)\otimes\mathrm{Hom}\left(B,D\right) \to \mathrm{Hom}\left(A\otimes B,C\otimes D\right)$,</p> <p>$f\otimes g\mapsto \left(a\otimes b\mapsto f\left(a\right)\otimes g\left(b\right)\right)$</p> <p>is injective.</p> <p>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?</p> <p>Also, is there any application of the above theorem outside of the two cases I mentioned?</p> <p>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?</p> http://mathoverflow.net/questions/72013/homa-c-homb-d-injects-into-homab-cd-when-why/72019#72019 Answer by Mahdi Majidi-Zolbanin for Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? Mahdi Majidi-Zolbanin 2011-08-03T18:07:51Z 2011-08-03T18:07:51Z <p>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 $D_1=0$. </p> http://mathoverflow.net/questions/72013/homa-c-homb-d-injects-into-homab-cd-when-why/72021#72021 Answer by a-fortiori for Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? a-fortiori 2011-08-03T18:10:06Z 2011-08-03T18:10:06Z <p>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 $i$.</p> http://mathoverflow.net/questions/72013/homa-c-homb-d-injects-into-homab-cd-when-why/72159#72159 Answer by Martin Brandenburg for Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? Martin Brandenburg 2011-08-05T09:04:31Z 2011-08-05T09:04:31Z <p>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)</p> <blockquote> <p><em>Lemma.</em> The natural map $V \otimes \prod_i W_i \to \prod_i (V \otimes W_i)$ is injective.</p> </blockquote> <p>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 <code>$((w_{ib})_{i})_{b} \mapsto ((w_{ib})_{b})_{i}$</code>, obviously injective.</p> <blockquote> <p><em>Corollary.</em> The natural map $\prod_i V_i \otimes \prod_j W_j \to \prod_{i,j} V_i \otimes W_j$ is injective.</p> </blockquote> <p>Proof: Apply the Lemma twice.</p> <blockquote> <p><em>Lemma.</em> The natural map $\hom(V',V) \otimes \hom(W',W) \to \hom(V' \otimes W',V \otimes W)$ is injective.</p> </blockquote> <p>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.</p>