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Let $V_i,W_i$ be finite dimensional vector spaces, for $i=1,2$. Assume we have homomorphisms $\phi_i:V_i\rightarrow W_i$.

Then, there is an induced map $\widehat{\phi_1 \times \phi_2} \in Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that $\widehat{\phi_1 \times \phi_2}(v_1 \otimes v_2) = \phi_1(v_1)\otimes\phi_2(v_2)$.

We can look at this construction as a map $\psi: Hom(V_1,W_1)\times Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$, where

$\psi:(\phi_1,\phi_2)\rightarrow \widehat{\phi_1 \times \phi_2}$.

It is easy to see that $\psi$ is a bilinear map. It follows (from the universal property) that we have an induced homomorphism: $\hat \psi:Hom(V_1,W_1)\otimes Hom(V_2,W_2)\rightarrow Hom(V_1 \otimes V_2,W_1 \otimes W_2)$ such that: $\hat\psi(\phi_1\otimes\phi_2)=\widehat{\phi_1 \times \phi_2}$.

My question: is $\hat\psi$ necessarily an isomorphism? Note that there is equality of dimensions (since both Hom,Tensor multiply them), hence clearly some isomorphism must exists between $Hom(V_1,W_1)\otimes Hom(V_2,W_2)\cong Hom(V_1 \otimes V_2,W_1 \otimes W_2)$.

If $\hat\psi$ is not isomorphism, is there some other "canonical" isomorphism?

I will just note that in the case of finite dimensional vector spaces, there is equality of dimensions, hence clearly some isomorphism must exists.

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    $\begingroup$ Take $W_1 = W_2 = A$, you can even assume it's a field then you're asking if $(V_1 \otimes V_2)^* \cong V_1^* \otimes V_2^*$. That's false if $V_1$ and $V_2$ are infinite dimensional. I can't even imagine how complicated things would become if $A$ is not a field, or not even a PID or anything... For $A$ a field my instinct tells me your $\hat\psi$ is always injective, but I'm not 100% sure. $\endgroup$ – Najib Idrissi Apr 28 '15 at 13:27
  • $\begingroup$ It need not be well defined. I have an example. $\endgroup$ – Matt Samuel Apr 28 '15 at 13:43
  • $\begingroup$ Najib: I agree. I have edited the question to focus upon the case of finite dimensional vector spaces. That was my original interest, I just thought maybe I missed some general theme which is not related to the specific fact that my objects were finite_dim vector spaces and not general modules. I was completely wrong there... $\endgroup$ – Asaf Shachar Apr 28 '15 at 13:43
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    $\begingroup$ $\widehat{\phi_1 \times \phi_2}$ is well defined via the universal property of tensor products. $\endgroup$ – Asaf Shachar Apr 28 '15 at 13:48
  • $\begingroup$ @egreg The expression $\phi_1(v_1) \otimes \phi_2(v_2)$ is linear in each variable in $v_1$ and $v_2$, if I'm not mistaken, so it defines a linear map $V_1 \otimes V_2 \to W_1 \otimes W_2$. And then this is linear in each of $\phi_1$ and $\phi_2$, hence the map $\hat\psi$. Am I missing something? It's just the bifunctoriality of $\otimes : \mathsf{Vect} \times \mathsf{Vect} \to \mathsf{Vect}$. $\endgroup$ – Najib Idrissi Apr 28 '15 at 14:04
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The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules

If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4).

Over a field, all modules are projective since they're free. So the answer is ‘yes’.

In the present case, here is a sketch of the proof:

Let $K$ be the base field. As $V_1\simeq K^m$ for some $m>0$ and similarly $V_2\simeq K^n$, and the $\operatorname{Hom}$ and $\,\otimes\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove: $$\widehat\Psi\colon\operatorname{Hom}(K,W_1)\otimes\operatorname{Hom}(K,W_2)\to\operatorname{Hom}(K\otimes K,W_1\otimes W_2)$$ is an isomorphism.

This results basically that for any vector space $V$ we have an isomorphism \begin{align*} \operatorname{Hom}_K(K,V)&\simeq V\\ \phi&\mapsto\phi(1) \end{align*} In detail, just consider the following commutative diagram:

enter image description here

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  • $\begingroup$ Thanks. Is there a quick way to see that $\hat\psi$ is an isomorphism in the particular case where all the modules are finite dimensional vector spaces? (I think this question should have a simple answer in terms of "undergraduate" linear algebra without the need to resorting to more "heavy" tools\theorems from abstract commutative algebra. $\endgroup$ – Asaf Shachar Apr 28 '15 at 13:57
  • $\begingroup$ @asaf shachar: I gave some details of a possible proof. See my completed answer. $\endgroup$ – Bernard Apr 28 '15 at 18:44
  • $\begingroup$ Excuse me... I want to ask that is this property holds for $> 2$ vector spaces? I think it works because the diagram listed above seems work for $> 2$ vector spaces. More generally, in module case, is this property holds for $> 2$ modules (with suitable requirements...)?? Thank you! $\endgroup$ – Peter Hu Sep 2 '15 at 10:54
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To show things like this in the case of finite dimensional vector spaces, you can always fallback on picking bases:

Let $v_i^1, \ldots, v_i^{m_i}$ be a basis of $V_i$ ($i=1,2$), and $w_i^1, \ldots, w_i^{n_i}$ be a basis of $W_i$ ($i=1,2$). Let $f^{jk}_i : V_i \to W_i$ be given on the basis by $f_i^{jk}(v_i^l) = \delta_{jl} w_i^k$. Then $\{v_1^p \otimes v_2^q\}$ is a basis of $V_1 \otimes V_2$, and $\{f_i^{jk}\}$ is a basis of $Hom(V_i,W_i)$. Your map sends $(f_1^{jk},f_2^{rs})$ to the linear transformation given on the basis by $\widehat{f_1^{jk} \times f_2^{rs}}(v_1^p \otimes v_2^q) = \delta_{jp} \delta_{rq} w_1^k \otimes w_2^s$.

Those maps clearly form a basis of $Hom(V_1 \otimes V_2, W_1 \otimes W_2)$ for the same reason the $f_i^{jk}$ are a basis of $Hom(V_i,W_i)$.

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