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Mac Lane once said that he didn't invent category theory to study categories, but to study natural transformations.

But googling around a bit, I haven't found out what natural transformations intuitively are about (only their definition).

I want to ask:

  1. What is the intuitive notion formalized by "natural transformation" in category theory, and (in the simplest possible terms) why does this formalization capture that intuition?

  2. Do category theorists in general still agree with MacLane's statement that natural transformations are the raison d'etre of category theory?

To clarify what I mean with an analogy:

  • A topology on a set $X$ formalizes the intuitive notion that each point $x\in X$ "touches" some subsets of $X$ but not others.

  • If a set $X$ has a group structure, this formalizes the intuitive notion that elements of $x$ can be seen as "invertible actions" that can be applied in succession.

  • Natural transformations (and their base concept Functor) formalizes the intuitive notion that ...

Arnaud D.
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user600670
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2 Answers2

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Natural transformations formalize the intuitive notion that a morphism $F(X) \to G(X)$ is defined "independently of $X$".

I think for newcomers to category theory a good example is the natural morphism $V\to V^{**}$ for a vector space $V$, as opposed to $V\to V^*$ (where $V^*$ is the dual of $V$, i.e. the vector space of linear forms on $V$ - for a fixed field $k$)

The usual morphism $V\to V^{**}$ is defined as follows : $x\mapsto (ev_x : l\mapsto l(x))$. In a sense, this definition makes no appeal to the specificness of $V$: it's only defined using composition and knowledge about what a vector space is. "We have made no choice" in defining it.

Comapre it to the usual ways to define morphisms $V\to V^*$. Often, what one does to show that they are isomorphic in finite dimensions is start with a basis of $V$ $(e_1,...,e_n)$ and define $e_i^*: V\to k$ to be the linear form assigning to a vector $v$ its $e_i$ coordinate; and finally define $V\to V^*$ by $e_i\mapsto e_i^*$. In this definition we have made a choice of a basis of $V$ and in a sense we have used the specificities of $V$ to define it.

This relates to the definition of a natural transformation in that the square that is required to commute for a natural transformation $\eta : F\implies G$, $\require{AMScd} \begin{CD} F(X) @>{\eta_X}>> G(X);\\ @VVV^{F(f)} @VVV^{G(f)} \\ F(Y) @>{\eta_Y}>> G(Y); \end{CD}$ means that, "as $X$ varies, $\eta_X$ varies along with it". You can make an analogy with topology by saying that this $\eta_X$ "varies continuously with $X$".

You can clearly see that this square detects the choice of a basis for my second example because the choice of a basis will not be "coherent" between two vector spaces $V,W$ and a morphism $f:V\to W$. But for the first morphism $V\to V^{**}$, since we have made no choice, this morphism will be "coherent" with any map $f:V\to W$.

Another nice example is the usual nautral isomorphism $2^X\to \mathcal{P}(X)$ for a set $X$ which is defined by $f\mapsto f^{-1}(\{1\})$. Once again you can see that this morphism can be defined without knowing anything about the specific $X$.

The intuition "it can be defined without knowing anything about $X$" can be seen as : we're not defining a map $F(X)\to G(X)$, really, what we're doing is defining a map $F(-)\to G(-)$ ; and if our definition was not "coherent between all the objects", there would be a noncommutative square that would detect this

Maxime Ramzi
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  • Thank you! This looks promising. Could you specify what exactly the functors $F$ and $G$ and the cateogires $X$ and $Y$ represent in the example with $V,V^$ and $V^{*}$? – user600670 Oct 21 '18 at 10:04
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    $X,Y$ aren't the categories, they're objects of the domain category ! $F$ is the identity functor, and $G$ is the bidual functor, $V\mapsto V^{}$ which acts on arrows as $f\mapsto (f^{} : l\mapsto (\varphi\mapsto l(\varphi\circ f)))$. For $V^$ it's trickier since it isn't a covariant functor, it's contravariant contrary to the identity functor so we have to adapt the notion of natural transformation between a covariant and a contravariant functor, but we can do it (and it becomes obvious why there can't be a nonzero natural transformation $V\to V^$) – Maxime Ramzi Oct 22 '18 at 07:41
  • I'm trying to see the injectivity of the isomorphism $x \mapsto ev_x$. could you give an idea? – John Mars Mar 11 '21 at 05:11
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    @JohnMars : this is a classical linear algebra fact : for any nonzero $x$, there is a linear form $f$ which doesn't vanish on $x$ – Maxime Ramzi Mar 11 '21 at 08:19
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    Regarding Ramzi's comment, the linear form $f$ that doesn't vanish on $x$ always exists when the space is finite-dimensional. In infinite dimensions, you would need the Hahn-Banach theorem, which relies on the axiom of choice. – Stultus sum May 01 '23 at 00:56
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I believe I have found a very simple picture to understand it. Firstly, we need to think what is the conceptual idea behind a functor? I quote John Baez to answer:

“Every sufficiently good analogy is yearning to become a functor.” -- John Baez

Now, what are natural transformations? Suppose we have two ideas $C$ and $D$, let us say we can make analogy between them in two ways by functors $F$ and $G$. A natural transformation is when, we have enough regularity to correspond between these two analogies between two ideas $A$ and $B$. I show a picture:

enter image description here

I have not drawn all element and morphism in each category , but took a few for illustration purpose. See how the object p,r and morphism q get mapped under functor $F$, and also see how the same is mapped under $G$.

Now, a natural transformation is simply a a nice analogy between analogies. The relation between $p,q,$ under the map of $F$ and under the map of $G$ is that, there exists a transformation $\eta_F$ and $\eta_G$ such that:

$$ \eta_y \circ F (f)= G(q) \circ \eta_x$$

tryst with freedom
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