Why is it okay to think of naturally isomorphic sets as being equal to each-other?

Question

Okay first of all here's what I understand of the two notions —

A natural Isomorphism between two sets is an isomorphism that can be constructed without any choices involved, like choosing a basis or an inner product or something of that nature to define the isomorphism.

Equality of 2 sets implies that the 2 sets contain the exact same elements.

My questions are particularly in regard to the natural isomorphism between a vector space ($V$) & its double dual ($(V^*)^*$) for finite dimensional vector spaces.

Question 1: We can prove that a vector space ($V$) is naturally isomorphic to its double dual double dual ($(V^*)^*$) but we cannot prove that they are equal, correct? Saying $V = (V^*)^*$ is an additional assertion that has nothing to do with the mathematics, right?

Question 2: If so what's the motivation for making this assertion? I understand that this allows us to think of vectors as (1,0) tensors but is there a deeper reason for doing this?

Question 3: Consider a real inner product space $(V, <.,.>)$ equipped with a special distinguished inner product, then a map $$ \phi : V \longrightarrow V^*$$ $$ \ \ : v \longrightarrow <., v>$$ then $\phi$ ought to be a natural isomorphism too, right? Does this mean that we can equate $V = V^*$ such that $v = \phi (v)$ & draw no distinction between vectors & covectors?

Answer 1

A natural Isomorphism between 2 sets is an isomorphism that can be constructed without any choices involved, like choosing a basis or an inner product or something of that nature to define the isomorphism.

I have no idea what this means. If one is talking about isomorphism between two sets, then it means bijection. For sets, there is no vector space structure and hence "choosing a basis" is nonsense.

We can prove that a vector space $V$ is naturally isomorphic to its double dual $V^{**}$

This is not always true. Consider the infinite dimensional (topological) vector spaces. (Also, "algebraic dual spaces" and "continuous dual spaces" are two different notions.)

but we cannot prove that they are equal, correct?

Yes, if "equal" is in the sense of sets.

If so what's the motivation for making this assertion?

You mean isomorphism between $V$ and $V^{**}$? Answers should be in

Why do we care about dual spaces?

What is the motivation/application of dual spaces and transposes?

Consider a real inner product space $(V, \langle.,.\rangle)$ equipped with a special distinguished inner product, ...

What is the "special distinguished inner product"? A related topic to your third question is called Riesz representation theorem.

Answer 2

A natural isomorphism is a special case of a natural transformation. More precisely, it is a natural transformation with an inverse. This is the correct technical definition. Since natural transformations are morphisms between functors, then a natural isomorphism is really between functors or "constructions" rather than between specific vector spaces or specific sets.

For instance, when $V$ is finite-dimensional the natural isomorphism $V \cong V^{**}$ is really between the identity functor and the double dual functor on the category of finite-dimensional vector spaces.

Saying that $V = V^{**}$ is a convenient "abuse of notation." If we are working in set-theoretic foundations it is not literally true. Justifying this common and useful abuse of notation is one of the benefits of the univalence axiom in homotopy type theory, which is a new alternative foundation for mathematics.

With regards to the canonical isomorphism $V \cong V^*$ when $V$ is an inner product space, this isomorphism cannot be natural in the usual sense since the identity functor is covariant whereas the duality functor is contravariant.

See this mathoverflow thread for a discussion of naturality which you might find helpful.