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The concept of natural transformation is often given with no motivation for it at all. It is a strange definition that is for some reason central in category theory. But why? This question tells that natural transformations are meant to capture defining a family of morphisms $F(X)\to G(X)$ "independently of $X$". But this doesn't explain why should we consider natural transformations as the morphisms between functors.

It would seem much more natural to define morphisms of functors as commutative squares of functors, since thats the usual way to define morphisms between morphisms in categories in general. So why are natural transformations defined different? It feels like an arbitrary thing to do..

More worryingly, since natural isomorphisms are considered "the" isomorphisms between functors, it means that we use this strange definition to decide which functors to treat as "the same". In other mathematical structures like algebraic structures or models of a theory, it's clear how to define isomorphisms: these should be bijections that preserve operations and relations in both directions. Is there some similar way to define natural transformations?

ZFCarla
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  • How do you motivate the fact that the "right" definition of homomorphism between sets $X,Y$ with an action of a monoid $M$ is a function $f : X \to Y$ such that $\forall x,,f(m.x)=m.f(x)$? – fosco Jun 11 '22 at 06:00
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    What is a commutative square of functors? The way I interpret that is, a diagram I draw where all 4 of the vertices allow me to plug in an argument, and I obtain a commutative square. That sounds a lot like what you said about natural transformations defining morphisms "independent of X." – A. Thomas Yerger Jun 11 '22 at 16:27
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    There's something confusing about your question. You write It would seem much more natural to define morphisms of functors as commutative squares of functors. But isn't that exactly how natural transformations are defined? – Lee Mosher Jun 11 '22 at 16:34
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    @LeeMosher No. If we have two functors $F,G : C\to D$, a commutative square would consist of functors $A: C\to C$ and $B: D\to D$ such that $GA=BF$. – ZFCarla Jun 11 '22 at 17:32
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    Given two vector spaces $V$ and $W$, when someone speaks about a linear transformation from $V$ to $W$, I expect some kind of arrow from $V$ to $W$. And given two functors $F$ and $G$, each from a category $C$ to a category $D$, when someone speaks about a natural transformation from $F$ to $G$ I expect some kind of arrow from $F$ to $G$. I don't see how that is captured by giving one arrow from $C$ to $C$ and another from $D$ to $D$. The actual definition instead does give a kind of arrow from $F$ to $G$. – Lee Mosher Jun 12 '22 at 02:29
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    @ZFCarla To complicate your definition even more, it seems restrictive to require $GA(x)$ to be literally equal to $BF(x)$ for all $x$. It would be much more natural (no pun intended) to allow the two to be isomorphic, in which case the concept of natural isomorphism appears again – Exit path Jun 13 '22 at 17:12

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Since $\bf Cat$ is cartesian closed, for any two categories $\mathbf X,\mathbf Y$ there is an exponential category $\bf Y^X$. First notice that the objects of $\bf Y^X$ correspond to the functors $\bf 1\to Y^X$, that by the exponential property are the functors $\bf X\to Y$; hence if you want a "natural" notion of morphisms between such functors, you should look at what the arrows in $\bf Y^X$ are.

Similarly, the arrows of $\bf Y^X$ correspond to the functors $\bf 2\to\bf{Y}^\bf{X}$, that are those $\mathbf{2}\times\mathbf X\to\mathbf Y $, and so those $\bf X\to Y^2$; as you see, the notion of a functor $\bf X\to Y^2$ is exactly the same as that of a natural transformation of functors $\bf X\to Y$.

Here $\bf 1$ has one object and no non-identity arrows, while $\bf 2$ has two objects and one non-identity arrow.

Ezio Greggio
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    This feels like it might be begging the question. Isn’t $\mathbf{Cat}$ Cartesian closed in a $2$-categorical sense? That means we’re assuming we already have a well-defined notion of natural transformation. On the other hand, you can say that defining natural transformations in the way we do makes nice things like this happen – Exit path Jun 13 '22 at 17:40
  • I think that the notion of natural transformation, that one already defined as you say, is not necessarily to be assumed as a morphism of functor. This latter notion makes sense if one thinks of the functors $\bf X\to Y$ as the objects of a category, and in particular, the exponential category $\bf Y^X$ happens to have exactly such functors as objects. – Ezio Greggio Jun 13 '22 at 18:07
  • But how are you defining that category? – Exit path Jun 13 '22 at 18:10
  • That category is defined by its universal property, so it is a confirmation that choosing the natural transformations as the morphisms in a functor category, in a certain sense, is "canonical". – Ezio Greggio Jun 13 '22 at 18:21
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    Yes but my point is that the universal property is a $2$-categorical one, is it not? So to even formulate what a universal property is, you already need to have decided on a notion of morphisms between functors – Exit path Jun 13 '22 at 20:45
  • I don't know anything about higher category theory. If you want a reference, I read about this in the example 7.16 of Awodey's Category Theory, that indeed is named "Transcendental deduction of natural transformations". Anyway the universal property can be clearly defined even without using higher categories, and without talking necessarily of morphisms of functors. – Ezio Greggio Jun 13 '22 at 20:49
  • I think you’ve convinced me, thanks. We don’t need the $2$-categorical structure here, the universal property holds without it – Exit path Jun 13 '22 at 22:18
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    @leibnewtz $\textbf{Cat}$ is cartesian closed as a 1-category and as a 2-category, with the same exponential objects. – Zhen Lin Jun 13 '22 at 22:31
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Not really sure.

There's a little motivation from polynomial functors

$$ [S \triangleright P](x) = \Sigma s\colon S. C(P(s), x) $$

A natural transformation here is

$$ [P] \Rightarrow [Q] = \forall x. [P](x) \rightarrow [Q](x) $$

With a naturality requirement.

But this is just an unravelled version of a map between polynomials.

$$ [S \triangleright P] \Rightarrow[T\triangleright Q] = (f\colon S \rightarrow T) \triangleright (Q \circ f \Rightarrow P)$$

And this directly is a structure preserving map.

There's probably a way to define functors and natural transformations nicely in terms of total functional (anafunctors) two sided discrete fibrations (profunctors).

Molly Stewart-Gallus
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I think Ezio Greggio's answer is great, I like to expand a little bit with an interpretation that helped me a lot to think about the natural transformation $\alpha:F\to G$ for two functors $F,G:X\to Y$ is equivalently a functor $\alpha:X\times 2\to Y$ as being a "homotopy" from $F$ to $G$ ($2=\{0\to 1\}$ plays the interval role here). So $\alpha(-,0)=F$, $\alpha(-,1)=G$, and $\alpha(x,0\to 1):F(x)\to G(x)$. Let $f:a\to b$ in $X$, we see $\alpha$ maps the commutative square in $X\times 2$:

$\require{AMScd}$ \begin{CD} (a,0) @> (f,1_0) >> (b,0) \\ @VV (1_a,0\to 1) V @VV (1_b,0\to 1) V\\ (a,1) @> (f,1_1) >> (b,1) \end{CD}

to commutative square in $Y$:

$\require{AMScd}$ \begin{CD} F(a) @> F(f) >> F(b) \\ @VV \alpha(a,0\to 1) V @VV \alpha(b,0\to 1) V\\ G(a) @> G(f) >> G(b) \end{CD}

We now arrives at the textbook definition of natural transformation.

Kaa1el
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The gadgetry of category theory fits together elegantly in multiple ways. That's what allows the other insightful but sophisticated answers to the OP's question. But there is also, surely, an entry-level answer.

Let's start by trying to fix an appealing definition of isomorphism between functors. The basic idea we want is that functors $F, G\colon \mathsf{C} \to \mathsf{D}$ come to the same (up to isomorphism) if the $F$ and $G$ images of any commutative diagram in $\mathsf{C}$ (or any portion of $\mathsf{C}$ more generally) are isomorphic in $\mathsf{D}$. And that idea leads quickly and naturally(!) to the usual definition of a natural isomorphism between $F$ and $G$.

Then we recall that we typically think of isomorphisms categorially as special cases of some wider class of morphisms -- they are the morphisms which have two-sided inverses. OK: so we'll want to think of natural isomorphisms between functors as special cases of ... Well, what?

Natural transformations naturally(!!) drop out as the answer. A natural transformation between functors $F, G \colon \mathsf{C} \to \mathsf{D}$ again sends an $F$-image of some diagram in $\mathsf{C}$ to its $G$-image in a way which respects some of the internal structure of the original -- perhaps now collapsing non-isomorphic objects together, but at least preserving composition of arrows.

Or at any rate, that's basically the entry-level story I tell in the relevant chapters of my draft-in-progress Category Theory II which you can download from https://www.logicmatters.net/categories. See if that helps ...

Ooops ...how did I miss this was an old question from a year back?

Peter Smith
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  • What is the definition for diagrams in C to be isomorphic to one another? Especially if we define the diagrams themselves to be functors (from a small category into C), then doesn't defining isomorphism between diagrams (commutative or otherwise) a priori require having defined a notion of (iso)morphism between functors? I don't understand and as a result am concerned it might be begging the question. – hasManyStupidQuestions Feb 24 '24 at 22:37
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    "Especially if we define the diagrams themselves to be functors" Well, don't start there :). It is more natural to think of diagrams in the first instance as suitable collections of objects and arrows. And we can say what it is for such diagrams-as-objects-and-arrows to be mapped to each other by suitable suites of isomorphisms without presupposing anything yet about functors (and hence can use the idea in explaining what it is for two functors to be naturally isomorphic without going round in circles). – Peter Smith Feb 25 '24 at 14:42
  • (It's not clear to me what definition of morphism between diagrams is implicitly invoked in this answer.) I do agree that if we can choose a definition of morphism between diagrams, from reasonable first principles that don't implicitly depend on the definition of natural transformation, then it answers the question. We may need to be careful though, because these collections of objects and arrows turn out to be equivalent to specifying certain classes of functors, one can not choose a definition of morphism between them without also implicitly choosing a definition of morphism between (1/2) – hasManyStupidQuestions Feb 25 '24 at 17:52
  • certain classes of functors. And so one could unintentionally sneak in/assume the definition of natural transformation while doing so. As a shameless plug for my own question, I think one can make your recipe work (for the diagram induced by the "walking arrow") using double categories https://math.stackexchange.com/questions/4870033/are-commutative-squares-in-some-sense-universal-among-edge-symmetric-double-ca although I'm not sure. As an aside, I really enjoy your notes on Category Theory and the notes about the Galois connection between syntax and semantics. They are very helpful. (2/2) – hasManyStupidQuestions Feb 25 '24 at 17:55