I'm trying to check that if $\mathfrak g$ is a Lie algebra and and $V,W,Z$ are $\mathfrak g$-modules, then there is an isomorphism $$ \text{Hom}_\mathfrak g (V\otimes W,Z) \cong \text{Hom}_\mathfrak g(V, \text{Hom}_k(W,Z)). \tag{$\star$} $$ where for vector spaces $V_1$ and $V_2$ the set $\text{Hom}_\mathfrak g (V_1,V_2)$ is the subset of $\text{Hom}_k (V_1,V_2)$ (notice the subindex is $k$ instead of $\mathfrak g$) that consists of linear maps compatible with the action of $\mathfrak g$ on $V_1$ and $V_2$. I thought I found a natural isomorphism between (notice the subindex is $k$ instead of $\mathfrak g$) $$ \text{Hom}_k (V,W) \cong \text{Hom}_k(V, \text{Hom}_k(W,Z)). $$ described as follows: define a map \begin{align*} L_1 : \text{Hom}_k ( V \otimes W , Z) &\longrightarrow \text{Hom}_k(V , \text{Hom}_k (W,Z))\\ T & \longmapsto \big\{ v \mapsto T(v \otimes \bullet)\big\}. \end{align*} and define a map \begin{align*} L_2: \text{Hom}_k(V , \text{Hom}_k (W,Z)) &\longrightarrow \text{Hom}_k ( V \otimes W , Z)\\ \Phi & \longmapsto \big\{ u \otimes w \mapsto \Phi(u)(w) \big\} \end{align*} These maps are clearly linear, and I think I've proven that they are inverse to each other.
This seems so natural that I can't help but think that this will restrict to the isomorphism desired in $(\star)$, but I haven't been able to show this. Is it false? Is my attempt misguided?
