How do natural transformations capture a “lack of arbitrarity”/uniqueness

Question

I've read natural transformations are the main motivation for category theory so I'm trying to get a handle on them, but having difficulty.

In this question and this quora question it is asked what the intuition behind natural transformations are. The answers seem to say natural transformations formalizes the notion a map has a "lack of arbitrarity" in its definition. However this is not clear to me for two reasons:

1. Aren't natural transformations always from categories to themselves? Whereas the notion of unique maps could be from one category to another.

2. Why does the commutative square suggest a "lack of arbitrarity"/uniqueness? I can see this might be the case, but I don't see why it should be the case.

Edit: To clarify my comments to FShrike, the dual is used as an example of non-naturality.

But if we consider a set of isomorphisms $\eta_{V}: V \rightarrow V^{*}$ for all $V$, then we can define the functor $F$ such that $F(V)$ is the codomain of $\eta_V$ and for $f:V \rightarrow W$, $F(f) = \eta_W f\eta_V^{-1}$. $F(f)$ are well defined since $\eta_{\_}$ are isomorphisms and
$$
F(f)\eta_V = \eta_Wf
$$
making $\eta_{\_}$ a natural transformation from $I$ to $F$ if there isn't a mistake.

Answer 1

$\newcommand{\hom}{\mathsf{Hom}}\newcommand{\fd}{\mathsf{FDVect_k}}$I’m not sure if it is the “main motivation” in modern terms, but as a historical note I believe Eilenberg and McLane first felt the need to define naturality, which prompted a definition of functor and then of category.

Anyway, consider a candidate class of arrows $\alpha_A:F(A)\to G(A)$ where $F,G$ are functors and $A$ objects in their common domain. Suppose $\alpha$ has some “arbitrary choice” in its definition, for example its definition relied on a particular element of $A$ (if $A$ is a set). Morally this is unnatural. Categorically, it is extremely unlikely that this arbitrary choice would stand up to inspection when you put it through the commutative square: for every arrow $f:A\to B$, for any other $B$, we demand that $\alpha_B\circ F(f)=G(f)\circ\alpha_A$. Not only that, but we demand this for all $A$ as well. Why is it extremely unlikely? Well, supposing for the sake of clarity that the objects are sets and the arbitrary choice involves some arbitrarily-chosen elements of these sets (for example, the isomorphism between $V$ and $V^\ast$, $V$ a finite dimensional vector space, is unnatural as we need to pick an arbitrary basis (=element of the set) to construct this isomorphism), then it would be very easy for me to construct a counterexample arrow $f:A\to B$ that played on this arbitrary choice, to make the square non-commutative, unless $F,G$ were extremely trivial or similar functors.

The naturality axiom also means that there is only one arrow you can construct as a composite (the diagonal of the square). If this didn’t hold, computations would be more difficult, but you’d also have to sit and worry about which route is the better one. This is unsatisfactory and an alarm bell that the approach you’re taking might not work, or might not be so meaningful, especially as you cannot make this decision in the abstract case. We should not have to make this decision, and there are distinct advantages for not needing to worry about this over needing to concern yourself with this (annoying) discrepancy.

I am new to category theory, but that is my take on it. Demanding the square commute always is A) exceptionally useful in proofs and computations, and B) it essentially removes the possibility of the arrows $\alpha_A$ having arbitrary choice (the space of counterexamples is potentially huge otherwise). We care about removing arbitrary choice in category theory and mathematics generally because if we can do something without it, that suggests a deeper link, and if a proof relies on an arbitrary choice you might say it is “unsatisfying” or not as meaningful. For example, you might have to worry if you made a bad choice, somehow.

Apropos your example natural isomorphism:

Let $k$ be any field and $\sf{FDVect_k}$ the category of finite dimensional $k$-vector spaces and the linear maps between them.

Yes, your arrows $(\eta_V)_{V\in\fd}:V\to V^\ast$ define a natural isomorphism $\eta:1_\fd\to F$. However, there is an annoying complication - $F$ itself is not "natural" and no-one would call it the "Dual functor". Why? Well,... its definition contains arbitrary choice :) This may feel very circular. To clarify, $V\simeq V^{\ast\ast}$ is considered a natural isomorphism since there is a natural isomorphism $1_\fd\to(\cdot)^{\ast\ast}$ where $(\cdot)^{\ast\ast}$ is the bidual functor, which has a standard, widely accepted and above-all natural definition. Category theory does not have language, afaik, to characterise the naturality of our own choices of standard functors here, but $(\cdot)^{\ast\ast}$ really is the only "reasonable" definition of a bidual functor. So, if $F$ is not "a good choice" of dual functor, what is?

More generally, fix $W\in\mathsf{FDVect_k}$. Define $\hom(\cdot,W):\fd^{\color{red}{\text{op}}}\to\fd$ by the object map: $$V\mapsto\hom(V,W):=\fd(V,W):=\{T:V\to W,\,T\text{ is a linear map}\}$$Because this new set $\hom(V,W)$ is itself a vector space with a canonical pointwise-defined linear structure. The arrow map is defined by: $$(f:V_1\to V_2)\mapsto(f^\ast:\hom(V_2,W)\to\hom(V_1,W))\\f^\ast(g)=g\circ f$$From this it is good to try and understand why $\hom$ needs to be defined on the dual category, the opposite category. It is hopefully clear that this is a natural way to map arrows, and that there isn't really any other way to do it.

If you set $W=k$ the one-dimensional vector space over itself, we define $(\cdot)^\ast:=\hom(\cdot,k)$ since $\hom(V,k)=V^\ast$. In this instance, you might recall from linear algebra that every matrix has a transpose. Actually, you may recall from dual spaces that the transpose of a linear map $T:V\to W$ is really defined as a map $T^t:W^\ast\to V^\ast$. Here, $T^t=(\cdot)^\ast(T)=\hom(\cdot,k)(T)$, a nice connection and hopefully an encouragement in the naturality of this dual functor $(\cdot)^\ast$. This is the dual functor, the standard definition used. Notice it is impossible for there to exist a natural isomorphism $\eta:1_\fd\to(\cdot)^\ast$ since they have different domains. And, given a choice between $(\cdot)^\ast$ or your (arbitrary) $F$, the majority (if not all) of mathematicians would say $(\cdot)^\ast$ is "the more natural choice". If we inspect how your $\eta$ would actually work, we see $F$'s construction needs a lot of axiom-of-choice business to make it defined, and this is quite non-constructive and is impossible to calculate anything with in general.

To emphasise, the fact that mathematicians say "$V$ is not naturally isomorphic to $V^\ast$" this is what they mean: $\eta:1_\fd\to(\cdot)^\ast$ does not exist. That does not mean there are no other natural isomorphisms involving $V$ and $V^\ast$, it just means the isomorphisms there are just not isomorphisms of anything we would actually consider.

The bidual functor is defined as $(\cdot)^\ast\circ(\cdot)^\ast$. It is a nice exercise to show that the famous natural isomorphism of $V$ and $V^{\ast\ast}$ really is a natural isomorphism $\alpha:1_\fd\to(\cdot)^{\ast\ast}$.

A nice exercise which might build your understanding (I enjoyed this exercise a lot, it gave me a lot of confidence that I understood the material - I'm a beginner too :)): (Source, Leinster's basic category theory):

Let $\mathscr{B}$ be the category of finite sets and bijections. Given $X\in\mathscr{B}$, let $\sf{Sym}(X)$ be the set of permutations on $X$ (the set of all bijections $\pi:X\to X$) and let $\sf{Ord}(X)$ be the set of all total orders on $X$ (which are essentially just strings $x_0\le x_1\le\cdots\le x_n$). Extend $\sf{Sym},\sf{Ord}$ to functors $\mathscr{B}\to\sf{Set}$ (i.e. provide an arrow map, an object map has already been given) in a natural way (it is important that $\mathscr{B}$ is equipped with bijections only whereas $\sf{Set}$ is not). Exhibit isomorphisms $\sf{Sym}(X)\simeq\sf{Ord}(X)$ for every finite set $X$, but show that it is impossible for there to exist a natural isomorphism (or indeed, any natural transformation at all!) $\alpha:\sf{Sym}\to\sf{Ord}$. Hint: "consider identity permutations".