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.
Best Answer
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: