Why is important to know if transformations between functors is natural or not?

Question

I'm studying category theory and I understand what a natural transformation is. I also understand that intuitively speaking it's like defining a polymorphic function that's independent of types, so to speak. However, my confusion is why do we want to know if a transformation is natural?

If a transformation between functors isn't natural then what can I not do and how does that differ from when they are natural?

It seems some aspect of composability may be impacted but there still will be a "path" $F: C \rightarrow D$ and it may not be equivalent in some sense to another $G: C \rightarrow D$. This is where my confusion lies:

1. When is this unnaturality possible? I can definitely conjure arbitrary/random functions that lead to unnaturality but all other (relevant) functions seem to be natural since $F,G$ both have the same source/destination I'm having a hard time fathoming when exactly would they not be natural.
2. Is there a "naturally occurring unnatural transformation" (pardon the wordplay) where $F,G: C \rightarrow D$ and what does it "prevent" us from concluding/proving/moving forward in the categorical framework?

I realize I may be getting a little ahead of myself here and maybe the answer will be clearer as I progress through the course but this question has been nagging away at me for a while and I'm not sure I understand the implications well enough to know why is necessary to have such a definition of natural transformation between functors?

Answer 1

When you have functors, you like to do a lot of stuff with them. For instance, if $F$ is a functor and $A$ has a group action $G$, then $F(A)$ inherits that group action.

If you have a natural isomorphism $F\cong H$, then this induces an isomorphism $F(A)\cong H(A)$ compatible with the respective $G$-actions.

This is because for all $g\in G$, the following diagram commutes, by naturality :

$\require{AMScd}\begin{CD}F(A)@>>> H(A) \\ @VF(g)VV @VH(g)VV \\ F(A) @>>> H(A)\end{CD}$

So for instance you can take fixed points (if those exist in your category) and get the same result : $F(A)^G\cong H(A)^G$, or orbits, $F(A)_G\cong H(A)_G$, etc.

Essentially, anything you can do to $F$, you can also do to $H$, and get results that are still isomorphic.

If the functors were nonnaturally isomorphic though, then maybe you can do the same thing to both and end up with different results. Or in the same vein, you can ask the same (categorical) question about both and end up with a different answer.

Examples of what can go wrong can be found here , I'll spell out a few to give examples with constructions and examples with questions :

1- Functors $BG\to \mathsf{Set}$ are nonaturally isomorphic if and only if their evaluation on the single object $*\in BG$ have the same cardinality. In particular, when you take their limit (i.e. the fixed points) you may get very different results. This gives you two functors that are pointwise isomorphic yet the same construction on both yields different results.

2- On finitely generated abelian groups, $\mathrm{Tor}_1^\mathbb Z(\mathbb Z/m,-)$ and $\mathbb Z/m\otimes-$ are pointwise isomorphic (because of the structure theorem), but they're not isomorphic as functors. In particular the former is left exact but not right exact, and the latter is right exact but not left exact.

3- A bit of both somehow : the universal coefficient exact sequence in cohomology is split, but nonnaturally so; in particular $H^n(-;\mathbb Z)$ is not naturally isomorphic to $\hom(H_n(-;\mathbb Z),\mathbb Z)\oplus \mathrm{Ext}^1_\mathbb Z(H_{n-1}(-;\mathbb Z),\mathbb Z)$. Concretely, there are some maps $X\to Y$ which induce $0$ on $\hom(H_n(-;\mathbb Z),\mathbb Z)\oplus \mathrm{Ext}^1_\mathbb Z(H_{n-1}(-;\mathbb Z),\mathbb Z)$ but not on $H^n(-;\mathbb Z)$

How can these happen ? Well in example 1-, this happens because all of the data of a functor $BG\to \mathsf{Set}$ is contained in how it acts on arrows, so if you forget that, you completely lose the information on your functor, and it becomes super easy for things to be isomorphic. That's what will happen most of the time : you're essentially asking about the difference between $Fun(C,D)$ and $Fun(Ob(C),D)$, and obviously $Ob(C)$ and $C$ are wildly different.

In examples 2- and 3-, the reason is that you have some decomposition or structure theorems that provide noncanonical decompositions (for 2- : the structure theorem for finitely generated abelian groups, for 3- the fact that a subgroup of a free abelian group is free abelian + the leftover choice in finding splittings of surjections to free abelian groups) which are themselves done objectwise (for instance, given two free abelian groups and a morphism between them, it need not respect any bases in any way, shape or form)

So to summarize :

we care about functors being naturally isomorphic because it means we can perform the same constructions on both and get the same results, and ask the same questions and get the same answers, no matter what they are; and also (perhaps this is more important) because the major part of the data of a functor is in how it acts on arrows (*). There are naturally occurring examples of nonnaturally isomorphic functors, coming for instance from noncanonical decomposition or structure theorems that do not respect the maps.

Note that (*) is not very well reflected in how we introduce functors in practice : they're often only defined on objects, leaving the action on arrows for the reader to infer.