# Natural Transformations Without Objects

So I've been thinking about the definition of categories as just arrows with a defined composition (i.e. without objects). I understand this is silly, but it's fun and I have a question about it: functors are easy to define in this setting; they're just maps that preserve the composition. But how does one discuss natural transformations in this setting, since we don't have objects around to "index" them by? I suppose we could index by identity arrows, but this feels wrong somehow, and I'm sure one of you smart people has a better answer.

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A natural transformation of functors $\mathbb{C} \to \mathbb{D}$ is the same thing as a functor $\mathbb{2} \times \mathbb{C} \to \mathbb{D}$, where $\mathbb{2} = \{ 0 \to 1 \}$. The domain of the natural transformation is the restriction to $\{ 0 \} \times \mathbb{C}$, and the codomain is the restriction to $\{ 1 \} \times \mathbb{D}$.

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How does composition of natural transformations work when viewed this way? – Shay Aug 1 '14 at 20:22
Use the category $\mathbb{3} = \{ 0 \to 1 \to 2 \}$. (Exercise.) – Zhen Lin Aug 1 '14 at 20:23
Oh my goodness, why did I never realize that a natural transformation looks like a... for lack of a better term, pre-homotopy? facepalm – Malice Vidrine Aug 1 '14 at 20:29
I like to think that is because the definition of homotopy is wrong! But unfortunately exponential objects in $\mathbf{Top}$ are a bit complicated to describe (assuming they even exist). – Zhen Lin Aug 1 '14 at 20:31
– Martin Brandenburg Aug 1 '14 at 20:59

Identity arrows are it. For an identity $x$ and functors $F,G$, the component of a natural transformation $\sigma:F\to G$ at $x$ is going to be a morphism for which $F(x)$ and $G(x)$ are left and right identities, respectively.

Keep in mind in the normal definition of a category, there's a bijection between objects and their identities; indexing with either one gives you essentially the same natural transformation even before we consider tossing out the objects.

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That is correct, but it is just an immediate reformulation (using the equivalence between the two definitions of a category). – Martin Brandenburg Aug 1 '14 at 21:01
True enough, and that was how I hoped it would be read. The way that the original poster thinks "feels wrong" works in a perfectly adequate and straightforward way. – Malice Vidrine Aug 1 '14 at 21:13
(Naturally, Zhen Lin's answer is much more interesting.) – Malice Vidrine Aug 1 '14 at 21:31