Tensor product group representations and spaces of intertwiners. Let $V_{1}$, $V_{2}$, $W_{1}$, and $W_{2}$ be the carrier spaces of  representations of some finite group $G$. Suppose also that $G$ acts trivially on $V_{1}$ and $V_{2}$. I would like to prove the following isomorphism between spaces of intertwining maps:
$$
\text{Hom}_{G}(V_{1}\otimes W_{1},
V_{2}\otimes W_{2})\cong
\text{Hom}(V_{1},V_{2})\otimes
\text{Hom}_{G}(W_{1},W_{2}).
$$
I think this is probably very easy, but I'm getting a little confused with the notation.  
Edit: I've attempted a proof below. Any comments would be appreciated.
 A: First note the following construction. Given two representations $(\pi,V)$ and $(\eta,W)$ of a group $G$, we can form the the homomorphism representation $(\rho_{V,W},\text{Hom}(V,W))$, defined by
$$
\rho_{V,W}(g)T:=\eta(g)\circ T\circ\pi(g^{-1}).
$$
The space $\text{Hom}_{G}(V,W)$ of intertwiners can be interpreted as the fixed points of $\text{Hom}(V,W)$ under this group action.
Now consider the isomorphism
$$
\psi:\text{Hom}(V_{1},V_{2})\otimes
\text{Hom}(W_{1},W_{2})\rightarrow
\text{Hom}(V_{1}\otimes W_{1},
V_{2}\otimes W_{2})
$$
obtained by sending each pure tensor $T\otimes S$ to the tensor product $T\otimes S$ and extending by linearity. The spaces $\text{Hom}(V_{1},V_{2})\otimes \text{Hom}(W_{1},W_{2})$ and $\text{Hom}(V_{1}\otimes W_{1},V_{2}\otimes W_{2})$ carry the $G$-representations $\rho_{V_{1},V_{2}}\otimes\rho_{W_{1},W_{2}}$ and $\rho_{V_{1}\otimes W_{1},V_{2}\otimes W_{2}}$ respectively. The map $\psi$ intertwines these two representations, and so they are equivalent. As a consequence, $\psi$ must restrict to an isomorphism between the $G$-fixed space of its domain and the $G$-fixed space of its image. The latter is just $\text{Hom}_{G}(V_{1}\otimes W_{1},V_{2}\otimes W_{2})$, and the former is equal to $\text{Hom}(V_{1},V_{2})\otimes\text{Hom}_{G}(W_{1},W_{2})$, since $\rho_{V_{1},V_{2}}$ acts trivially on $\text{Hom}(V_{1},V_{2})$.
