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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.

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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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  • $\begingroup$ I’d just say the use of “bi”functor instead of “functor”, which is all these things are, could be confusing. One doesn’t talk about a bilinear map $A\otimes B\to C$, after all. $\endgroup$ – Kevin Carlson Oct 9 '18 at 15:57
  • $\begingroup$ @KevinCarlson The term is used in e.g. CWM. Personally I don't think there is much chance on confusion (so will not change). If I am wrong after all then your comment is probably enough. $\endgroup$ – drhab Oct 9 '18 at 17:20
  • $\begingroup$ I don't really think the mention of a bi-functor is too confusing here. I had to look it up, but the real hardship in Riehl's book is the struggle to understand the enormous amount of examples from advanced maths that I have no experience with. Are most mathematicians really that well versed is such a wide variety of subjects? It's a scary thought. $\endgroup$ – j4nd3r53n Oct 10 '18 at 8:05
  • $\begingroup$ "...most..." I don't think so! At least I am not one of them. Guys like Riehl might be exceptions on that (or have well informed assistents). Do not despair. A lot of what you do not understand can just be skipped. If you are surrounded by a lot of incomprehensible math, then also you can try to work from top to bottom. Persevere! $\endgroup$ – drhab Oct 10 '18 at 8:33

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