$G$-representations, $W \otimes V^* \to \text{Hom}(V,W)$ Let $V$ and $W$ be finite-dimensional vector spaces. I know how to construct an explicit isomorphism of vector spaces $W \otimes V^* \to \text{Hom}(V,W)$ and show that it's an isomorphism. But if I supposed that $V$ and $W$ are $G$-representations of some group $G$, how do I show that the isomorphism above is an isomorphism of $G$-representations?
 A: Let $\rho_1: G \to GL(V)$ be a $G$-representation and $\rho_2: G \to GL(W)$ as well. We can make $V^*$ a $G$-representation $\rho_1^*: G \to GL(V^*)$ given by$$(\rho_1^*(g)(\phi))(v) = \phi(g^{-1}v).$$Then $W \otimes V^*$ is a $G$-representation $\rho_T: G \to GL(W \otimes V^*)$ by $$\rho_T(g)(w \otimes \phi) = (\rho_2(g)(w)) \otimes (\rho_1^*(g)(\phi)).$$In addition, we have $\text{Hom}(V, W)$ is a $G$-representation $\rho_H: G \to GL(\text{Hom}(V, W))$ given by $($for $T \in \text{Hom}(V, W)$$)$$$(\rho_H(g)(T))(v) = \rho_2(g)(T(\rho_1(g^{-1})(v))).$$As all the $\rho$'s are getting unwieldy at this point, we will just write the $G$ action as $g \cdot$ and hope everything is clear from context. So we can now see that the map $L$ from above is $G$-linear. For$$L(g \cdot (w \otimes \phi))(v) = L((g \cdot w) \otimes (g \cdot \phi))(v) = (g \cdot \phi)(v)g \cdot w$$and $$(g \cdot L(w \otimes \phi))(v) = g \cdot (L(w \otimes \phi)(g^{-1} \cdot v)) = g \cdot (\phi(g^{-1} \cdot v)w) = \phi(g^{-1} \cdot v)g \cdot w.$$And those are equal from how one defines the representation $V^*$.
