Natural Isomorphism $(V\otimes W)^*\cong V^*\otimes W^*$ According to Wikipedia, there exists a natural isomorphism $(V\otimes W)^*\cong V^*\otimes W^*$ for finite-dimensional vector spaces $V$ and $W$. I can show that given bases $v_i$, $i\in I$ and $w_j$, $j\in J$ with dual bases $v^*_i$, $i\in I$ and $w^*_j$, $j\in J$, an isomorphism is given by
$$
(V\otimes W)^*\ni D \mapsto \sum_{I\times J} v^*\otimes w^* D(v\otimes w).
$$
 I feel like the choice of bases above makes this isomorphism not natural. However, I am really bad at understanding "naturality".
 A: This is actually a bit tricky. There are two functors $F,G : \mathsf{Vect}^{op} \times \mathsf{Vect}^{op} \to \mathsf{Vect}$ given respectively by
$$F(V,W) = V^* \otimes W^*, \quad G(V,W) = (V \otimes W)^*$$
(and correspondingly on linear maps). Then there is a natural transformation $\varphi : F \to G$, in components:
$$\varphi_{V,W} : V^* \otimes W^* \to (V \otimes W)^*,$$
given by $\varphi(\alpha,\beta) : v \otimes w \mapsto \alpha(v) \beta(w)$. In other words, if $f : V \to V'$ and $g : W \to W'$ are linear map,
$$(f \otimes g)^* \circ \varphi_{V', W'}  = \varphi_{V,W} \circ (f^* \otimes g^*).$$
This map $\varphi_{V,W}$ is always injective; thus, when both $V$ and $W$ are finite-dimensional, it's an isomorphism (because of dimension considerations: $\dim(V^* \otimes W^*) = \dim(V) \dim(W) = \dim((V \otimes W)^*)$). If both are infinite-dimensional, then $\varphi_{V,W}$ is not an isomorphism.
To explicitly compute the inverse of $\varphi_{V,W}$ when both are finite-dimensional, you need to choose bases of $V$ and $W$ (and the inverse is indeed the map you have constructed). This is kinda weird, because we know the inverse exists without talking about bases, but it doesn't seem possible to actually write it down without bases.
