# Natural transformations correspond bijectively to functors … - but how?

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:

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.

• – Arnaud D. Oct 9 '18 at 11:54

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 $$\alpha_c:=H(c,\iota)$$.

• 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. – Kevin Carlson Oct 9 '18 at 15:57
• @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. – drhab Oct 9 '18 at 17:20
• 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. – j4nd3r53n Oct 10 '18 at 8:05
• "...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! – drhab Oct 10 '18 at 8:33