Natural Transformation and the freedom from artificial choices I'm reading Mac Lane's "Categories for the Working Mathematician"
On section $I.4$ on Natural Transformation, I've ran into into a statement I could not wrap my head around.
Given an abelian group G, we denote G's character group $D G$.
For a morphism of abelian groups, $f:G\rightarrow H$, we define $Df:DH \rightarrow DG$ as following:
$$
(Df)t = tf : t \in DH
$$
The writer then defines a transformation $\tau$ from an abelian group to it's double-character group. For each abelian group G, we define $ \tau_G :G \rightarrow D(DG) $ as
$$
(\tau_G g)(t) = tg : t \in DG, g \in G
$$
and shows that it satisfy the definition for natural transformations. So far, I was able to follow.
Mac Lane concludes:

One verifies at once that $\tau$ is a natural transformation. This
statement is just the expression for the elementary observation that
the definition of $\tau$ depends on no artificial choice of bases,
generators or the like.

While I can see that the provided transformation is not dependent on any choice of "bases,generators or the like", I can't really see how the fact that the transformation is natural formalizes this observation.
Any help?
 A: Recall the definition of a natural transformation: Let $F,G : C \to D$. Then an assignment of $\alpha_c : F c \to G c$, for each $c\in C$, is natural if, for every $f : c \to c'$ in $C$, we have that the following commutes:
$$
\begin{array}
(Fc            & \xrightarrow{\alpha_c} & G c \\\
\downarrow{Ff} &              & \downarrow{Gf} \\\
Fc'            & \xrightarrow{\alpha_c'}     & Gc'
\end{array}
$$
TLDR: In nice categories, the naturality condition requires that any choice made to define $\alpha_c$ and $\alpha_{c'}$ is preserved by $Ff$ and $Gf$, for every $f: c \to c'$. This does not happen when we make arbitary (incompatible) choices for $c$ and $c'$. See end remarks for a full answer to your question.
Now, suppose, for example, that $C=D=Vect_\mathbb{K}$, and $F=G=Id_{Vect_\mathbb{K}}$. There is of course the identity natural isomorphism, with each $\alpha_c$ the identity linear transformation. Now, what would happen if we try to define another natural isomorphism by making an arbitrary choice of basis in order to define any $\alpha_c$? Say the basis we choose for $Fc=c$ is $\{e_1,...,e_n\}$ and we defined $\alpha_c$ by sending this basis to another basis $\{v_1,...,v_n\}$. But, for $Fc'=c'$ we might have made the choice of basis of $\{e'_1,...,e'_m\}$ and defined $\alpha_c'$ by sending this basis to another basis $\{w_1,...,w_m\}$. Then each component is indeed an isomorphism since it is defined by a bijection of basis.
Now, what would the naturality condition would require? We would need the following diagram to commute:
$$
\begin{array}
(c            & \xrightarrow{\alpha_c} & c \\\
\downarrow{f} &              & \downarrow{f} \\\
c'            & \xrightarrow{\alpha_c'}     & c'
\end{array}
$$
Chasing the basis of $c$ along the top-right route lends $\{f(v_1),...,f(v_n)\}$. Chasing the basis of $c$ along the left-bottom route lends $\{\alpha_{c'}(f(e_1)),...,\alpha_{c'}(f(e_n))\}$. So naturality would require that $f(v_i)=\alpha_{c'}(f(e_i))$, for ever linear transformation $f:c\to c'$.
But there is no reason to expect this to happen for every $c, c'$ and every linear transformation $f$, since choice of basis is not necessarily preserved by an arbitrary linear transformation. In fact, I leave it as a challenge (it won't be that hard) to find specific $c, c'$, choice of basis, and $f$ where the naturality condition fails.
Now recall the identity natural transformation. This is indeed a natural transformation since the choice of $\alpha_c=1_c$ makes no reference choice of basis. So we do not have to worry about odd choices being preserved.
For another example, see Emily Rhiel's "Category Theory in Context". This book is free on here website, so I absolutely recommend getting it and taking a peak at it since it does present things differently that MacLane. The section on naturality has a good example.
Basically, this commutativity condition requires that any choice we make in defining $\alpha_c$ and $\alpha_{c'}$ is preserved by $Ff$ and $Gf$, for every $f : c \to c'$.
I should add, though, that natural transformations can take on a very abstract and more general nature than the examples that motive the "freedom form choices" intuition. E.g., in posets or in more abstract categories. In these cases, it is sometimes hard to keep the original intuition. So, the naturality condition does not in general express the idea of "freedom form choices", but rather it puts a coherence condition on the assignment of objects to morphisms. This is similar to how a functor has coherence conditions on the assignment of morphisms to morphisms (the composition laws). So the full answer to your question is this: the naturality condition does not in and of itself assert a freedom of choices condition. This only makes sense in certain concrete categories. The naturality condition just requires a coherence between the choice of morphism. That this "coherence" idea is the proper way to think of the naturality condition is justified by its fitting into a much larger hierarchy. For more, see this and this.
