What intuitive notion is formalized by "natural transformation" in category theory?

Solution 1:

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