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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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2 Answers 2

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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 at 20:22
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Use the category $\mathbb{3} = \{ 0 \to 1 \to 2 \}$. (Exercise.) –  Zhen Lin Aug 1 at 20:23
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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 at 20:29
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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 at 20:31
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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 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 at 21:13
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(Naturally, Zhen Lin's answer is much more interesting.) –  Malice Vidrine Aug 1 at 21:31

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