TITLE: Finite-dimensional space naturally isomorphic to its double dual?
QUESTION [11 upvotes]: The example that a finite vector space is naturally isomorphic to its double dual seems to be the canonical example of natural isomorphisms.
Concretely, there are two functors $\mathsf{Id}, {-^*}^* : \mathsf{FDVect} \to \mathsf{FDVect}$ which represent the identity functor and the double dual functor, and one can construct a natural transformation $\alpha : \mathsf{Id} \Rightarrow {-^*}^*$ as follows:
$$ \alpha_V(v)(f) = f(v)$$
However, in order to show that $\alpha$ is a natural isomorphism, one needs to find an inverse $\alpha^{-1}$.
I know that each $\alpha_V$ has an inverse, as the vector spaces are finite dimensional. I also know that if $\alpha$ is a natural transformation, then I can construct an inverse natural transformation by using the $\alpha_V^{-1}$'s.
However, the fact that $\alpha_V$ has an inverse $\alpha_V^{-1}$ does not seem very "natural" to me. I had to use a base of $V$ in order to construct $\alpha_V^{-1}$, and people always say that "choosing a base is not natural" (at least that is the reason why $V \cong V^*$ is not natural, right?). Despite of that, I know that $\alpha^{-1}$ is natural.
What is going on? Am I choosing a wrong $\alpha^{-1}$? Choosing a basis like that for $\alpha_V^{-1}$ is natural?

REPLY [10 votes]: You are not required to exhibit an inverse, only to prove existence. Thus, in the finite case, where the always-injective map to the second dual is actually a bijection, by dimension-counting, you can prove existence of an inverse without choosing anything: given $w$ in the second dual, there is unique $v\in V$ mapping to $w$. "Defined" $\alpha^{-1}(w)=v$.
It is true that defining "dimension" involves showing that the cardinalities of any two bases are the same, ... but it is not always the case that things need to be defined via a basis and then shown independent of the basis.