Natural isomorphism between a vector space and its double dual

Question

The usual way of showing the natural isomorphism between a vector space and its double dual is to prove that the map from $V$ to $V^{**}$ is injective and linear. And since $\dim V = \dim V^*$ and $\dim V^*=\dim V^{**}$, it follows that $\dim V = \dim V^{**}$, which establishes the bijection. I read that showing this isomorphism doesn't require us to choose a basis.

But I've also read that the proof of $\dim V^*=\dim V^{**}$ is similar to $\dim V = \dim V^*$, and proving the latter does require us to choose a basis. So won't establishing that $\dim V^*=\dim V^{**}$ (and hence, in effect, that $\dim V = \dim V^{**}$) also require a choice of basis? That way the proof of the isomorphism requires us to choose a basis at one point, so why is it called a natural isomorphism? Is there supposed to be a basis-free way of showing that $\dim V = \dim V^{**}$?

Answer 1

I read that showing this isomorphism doesn't require us to choose a basis.

This sentence is a little bit vague. What you probably read is that defining the isomorphism doesn't require a choice of basis. You can always define $\phi : V \to V^{**}$ by $\phi(v)(f) = f(v)$.

However, you cannot prove that it is a bijection without involving a basis somehow. Since $\phi$ is (in general) only an isomorphism when $V$ is finite-dimensional, you need to use this fact in your argument somewhere, and the dimensionality of a vector space is ultimately defined in terms of the cardinality of a basis for it.

Answer 2

why is it called a natural isomorphism?

In the category of finite dimensional vector spaces, there is a natural isomorphism of the identity functor to the double-dual functor. The resulting isomorphism for each object in the category is called "natural" because it is a component of this natural isomorphism of functors. (See Leinster, Tom. Basic category theory. Vol. 143. Cambridge University Press, 2014 example 1.3.14)

Actually, I need to reframe what I'm saying above. What is really the case is that mathematicians were aware of such things long before category theory, and category theory was invented in part to explain this in a precise way. So the explanation above is cheating a little bit by skipping over the story up to that point :) I recommend reading this post which explains the situation better!

I don't rightly know if you can say "you don't need to choose a basis" to prove it, but you can certainly show that the result is "independent of basis" meaning that any selection is as good as another and the selection itself is immaterial. Said another way, the isomorphism you come up with is the same no matter which basis you select.