Commuting of Hom and Tensor Product functors? 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.
 A: 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:

A: 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 $Hom(V_i,W_i)$.
