TITLE: Natural isomorphism in linear algebra: is the naturalness asymmetric?
QUESTION [8 upvotes]: Let $V$ be a finite-dimensional vector space. It is well known that there is a natural isomorphism between $V$ and its double dual $V^{\ast\ast}$ defined by $T(x)(f)=f(x)$ for every $x\in V$ and $f\in V^\ast$.
However, I am unable to write down a definition of $T^{-1}$ directly without any reference to $T$ and without picking any basis. Thus it seems to me that the isomorphism between the two vector spaces is not really so natural if we look at the direction from $V^{\ast\ast}$ to $V$.
Is it possible to define $T^{-1}$ in a direct and natural way? If not, does this asymmetry in the easiness of definition have any significance in linear algebra or other branches of mathematics?

REPLY [4 votes]: It is a common fallacy to attach to mathematical concepts meanings that pertain to the meaning of their name in the natural language.
For instance, a number can be irrational, but this doesn't mean it is not logic or not reasonable (the meanings I get from a dictionary); “real”, “imaginary” and “transcendental” are other examples. Don't attach to the adjectives used for classifying these numbers more than the mathematical definition says.
The name “natural isomorphism” was initially chosen based on examples from algebraic topology and the similarity with the already known case of the double dual was noticed, where the isomorphism $V\to V^{**}$ (for a finite dimensional vector space $V$) can be defined in “a natural way”.
When other instances of “natural (iso)morphisms” came around, it became apparent that a precise definition had to be found, not based on the intuition that the “definition is easy and obvious”, because in some cases it was definitely not easy and obvious, although it fitted the framework. And category theory was born.
If $F,G\colon \mathbf{A}\to\mathbf{B}$ are (covariant) functors, a natural morphism $\eta\colon F\to G$ is a family of morphisms $\eta_X\colon F(X)\to G(X)$, one for each object $X$ in $\mathbf{A}$ in such a way that, given any morphism $f\colon X\to Y$ in $\mathbf{A}$, we have the commutative diagram
$$\require{AMScd}
\begin{CD}
F(X) @>\eta_X>> G(X) \\
@V{F(f)}VV @VV{G(f)}V \\
G(Y) @>\eta_Y>> G(Y)
\end{CD}
$$
that is, $\eta_Y\circ F(f)=G(f)\circ\eta_X$. If each $\eta_X$ is an isomorphism, we say that $\eta$ is a natural isomorphism.
It easily follows from the definition that if $\eta$ is a natural isomorphism $\eta\colon F\to G$, then the collection $(\eta_X^{-1})$ defines a natural isomorphism $\eta^{-1}\colon G\to F$.
In the particular case when $\mathbf{A}=\mathbf{B}$ is the category of finite dimensional vector spaces over some fixed field, $F$ is the identity functor and $G$ the double-dual functor, then there exists a natural isomorphism $F\to G$, which is the well-known one.
Thus also the inverse is a natural isomorphism. But nothing in the definition suggests that this inverse can be “computed in a natural way”. And indeed it cannot: if it could, then the “natural” definition would also apply to infinite dimensional spaces, but this is not possible.