Let $M$ be a smooth manifold. Let $(T_pM)_{alg}$ denotes the "algebraist's" tangent space at $p\in M$, that is the tangent space via derivations, and $(T_pM)_{kin}$ denote the "kinematic" or "geometer's" tangent space at $p$, that is the tangent space via curves.
We can define a map $\Phi:(T_pM)_{kin}\to (T_pM)_{alg}$ as $$\Phi([\gamma])=(f\mapsto d(f\circ \gamma)/dt|_{t=0}) \text{ for all } f\in \mathcal C^{\infty}(M)$$ for all $[\gamma]\in (T_pM)_{kin}$
It can be shown that this map is well defined and is an isomorphism.
Now here is my question. I recently came across the concept of a natural isomorphism. So I was wondering whether the isomorphsim above is a natural one?
I have read this at many places that an isomorphism between vector spaces is natural if it doesn't depend on the choice of basis. Here our map $\Phi$ is not dependent on a basis. So we would say that it is a natural isomorphism. I find this rather vague.
But how do we say it precisely in category theoretic terms? So far I only know about natural isomorphism between two functors, both taking objects of a category $C$ to a category $D$. I haven't seen a precise definition of a natural isomorphism between two vector spaces.
Can somebody please enlighten me on this?
The intuitive definition is that the two objects are the same in every way, except in the notation in which you write it. For example, suppose $V$ is a vector space over $\mathbb{R}$, say finite dimensional, with a non-degenerate inner product. It is an easy exercise to see that $V$ and $V^*$ ($=\text{hom}(V,\mathbb{R})$) are naturally isomorphic. The way you do it, is send $v\in V$ to $v^*$, where $v^*$ is defined as the linear map $x\mapsto \left< v,x\right>$. Therefore, the elements in $V$ "look like" $v$, while the elements in $V^*$ "look like" $v^*$. So we say they are "naturally isomorphic". It is common for people to abuse notation and write $V=V^*$.
An example when the isomorphism is not natural, is say that $V$ is just a vector space, then it is still true that $V\simeq V^*$, as their dimensions are equal. However, the isomorphism is not natural.
A more precise meaning is in terms of preserving a commutative diagram. If you have a bunch of natural maps, fitting together in a diagram, it will be commutative. To clarify consider the following example,
Let $f:V\to V^*$ be an isomorphism, and $g:V^* \to V^{**}$ be another isomorphism, and $h:V\to V^{**}$ be a third isomorphism. It is not necessarily true that $h = gf$. However, if $V$ has inner product and $f$, $g$ are natural isomorphisms, and $h$ is the natural isomorphism of $V$ with its bidual then the diagram commutes.