Natural transformations correspond bijectively to functors … – but how

category-theory

Lemma 1.5.1 in Emily Riehl's book 'Category Theory in Context' states:

Given a pair of functors $F,G \colon C \rightrightarrows D$, natural transformations corrspond bijectively to functors $H \colon C \times \mathbb{2} \rightarrow D$ so that the following commutes:

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This is apparently obvious, but I keep going in circles around it. Naturality means that given $x \xrightarrow{f} y$, there are arrows $\alpha \in D$, so $\alpha_y \circ Ff = Gf \circ \alpha_x$, and the proof should construct $H$ from $\alpha$ and vice versa. I just can't get my thinking into the right shape.

Best Answer

Let $\alpha:F\stackrel{\bullet}{\to}G$ denote a natural transformation.

Let $\iota:0\to 1$ denote the unique arrow $0\to 1$ in category $\mathbf2$.


Now define $H:\mathcal C\times \mathbf2\to\mathcal D$ by stating that $H(-,0)=F$, $H(-,1)=G$ on objects and arrows in $\mathcal C$. Further for every object $c$ of $\mathcal C$ let $H(c,\iota):=\alpha_c:F(c)\to G(c)$.

Then $H$ can be shown to be a bifunctor $\mathcal C\times \mathbf2\to\mathcal D$.


If conversely $H$ is a bifunctor $\mathcal C\times \mathbf2\to\mathcal D$ then it induces a natural transformation $\alpha:F\stackrel{\bullet}{\to}G$ defined by $\alpha_c:=H(c,\iota)$.

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