I was reading this question with the same title at MathOverflow, which defines natural transformations in the following way:

given two functors $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ a natural transformation is a functor $\varphi \colon \mathcal C \times 2 \to \mathcal D$, where $2$ is the arrow category $0 \to 1$, such that $\varphi(-,0)=\mathcal F$ and $\varphi(-,1)=\mathcal G$.

This relates $\varphi$ to $\mathcal F$ and $\mathcal G$ on objects, but what about arrows? Why don't we also need to specify that $\varphi(-, id_0) = \mathcal F$ and $\varphi(-, id_1) = \mathcal G$?


1 Answer 1


There are two directions to this proof.

One direction is that given a functor $\varphi: \mathcal C \times 2 \to \mathcal D$, there is a corresponding natural transformation $\varphi(-, 0) \to \varphi(-, 1)$. $\varphi(-, 0)$ is a whole functor $\mathcal C \to \mathcal D$. The action on objects is obvious (simply evaluate $\varphi$ at the pair $(c, 0)$. If you haven't seen this before, the action on morphisms might not be obvious. Morphisms in $\mathcal C \times 2$ are defined to be pairs of morphisms in $\mathcal C$ and $2$, so a priori, $\varphi(f, 0)$ doesn't make any sense. However, it's typical with functors of multiple variables that an object is also shorthand for the identity at that object. That is, $\varphi(f, 0)$ is $\varphi(f, id_0): \varphi(c, 0) \to \varphi(c', 0)$.

Then, the natural transformation $\varphi(-, 0) \to \varphi(-, 1)$ is simply $\alpha_c := \varphi(c, \to)$, where $\to$ is the unique arrow $0 \to 1$ in $2$.

The other direction is that given a natural transformation $\alpha: \mathcal F \to \mathcal G$, there is a corresponding functor $\varphi: \mathcal C \times 2 \to \mathcal D$ such that $\varphi(-, 0) = \mathcal F$ and $\varphi(-, 1) = \mathcal G$. The behavior of $\varphi$ on objects is determined by the conditions that it's equal to the given functors at $0$ and $1$. For example, $\varphi(c, 0) = \mathcal F(c)$.

That leaves the action of $\varphi$ on morphisms. $\varphi(f, \to): \varphi(c, 0) \to \varphi(c', 1)$, i.e. $\mathcal F(c) \to \mathcal G(c')$. The natural choice is then the diagonal of the commutative diagram

$$ \require{AMScd} \begin{CD} \mathcal F(c) @>{\mathcal F(f)}>> \mathcal F(c')\\ @V{\alpha_c}VV @VV{\alpha_{c'}}V \\ \mathcal G(c) @>>{\mathcal G(f)}> \mathcal G(c') \end{CD} $$

Finally, one should really show that going one direction then the other leaves you where you left off. Once functorality of $\varphi$ and naturality of $\alpha$ are proven, that gives a bijection between functors of that certain form and natural transformations.

  • $\begingroup$ Thanks! That is indeed what was confusing me. Though now I see there's something else I may have been overlooking. I guess we don't need to explicitly specify $\varphi(id_c, \to)$; they're just unspecified choices that make $\varphi$ functorial (i.e., make the diagram commute). $\endgroup$
    – A_P
    Dec 21, 2019 at 5:24
  • $\begingroup$ $\varphi(id_c, \to)$ is a special case of $\varphi(f, \to)$. In particular, it ends up just being $\mathcal G (id_c) \circ \alpha_c = \alpha_c \circ \mathcal F (id_c) = \alpha_c$. $\endgroup$
    – SCappella
    Dec 21, 2019 at 10:35
  • $\begingroup$ I'm confused. If we start with an $\alpha$, then we have to explicitly connect $\varphi$ to it, right? So we simply set $\varphi(id_c, \to) = \alpha_c$ and we should be done? It being a special case of $\varphi(f, \to)$ just tells me then that $\mathcal G(id_c)\circ\varphi(id_c,\to)=\varphi(id_c,\to)\circ\mathcal F(id_c)=\varphi(id_c,\to)=\alpha_c$, no? (I called it an "unspecified choice" because in the original description we made no reference to an $\alpha$, since we were defining "natural transformation" in the first place. We just assumed such choices existed, I guess.) $\endgroup$
    – A_P
    Dec 21, 2019 at 16:26
  • $\begingroup$ @A_P Yes, there are two directions to this proof. For one, we start with $\varphi$ and define $\alpha_c := \varphi(id_c, \to)$. For the other direction, we start with $\alpha$. The behavior of $\varphi(-, 0)$ and $\varphi(-, 1)$ is determined by $\mathcal F$ and $\mathcal G$, but we have to use $\alpha$ to define $\varphi(f, \to)$ for morphisms $f$ in $\mathcal C$. $\endgroup$
    – SCappella
    Dec 21, 2019 at 19:00
  • $\begingroup$ Okay, I see where my confusion came from. In the second direction, one option is to define $\varphi(id_c, \to)=\alpha_c$ and get $\varphi(f, \to)$ for free by functoriality. The other is to instead define $\varphi(f, \to)$ as the diagonal and get $\varphi(id_c, \to)$ for free, as you've done. This second way somehow confused me. Thanks for the help! $\endgroup$
    – A_P
    Dec 21, 2019 at 20:54

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