# Natural transformations as categorical homotopies

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

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.

• 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). – A_P Dec 21 '19 at 5:24
• $\varphi(id_c, \to)$ is a special case of $\varphi(f, \to)$. In particular, it ends up just being $\mathcal G (id_c) \circ \alpha_c = \alpha_c \circ \mathcal F (id_c) = \alpha_c$. – SCappella Dec 21 '19 at 10:35
• 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.) – A_P Dec 21 '19 at 16:26
• @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 $\mathcal C$. – SCappella Dec 21 '19 at 19:00
• 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! – A_P Dec 21 '19 at 20:54