Extranatural transformations are a special case of dinatural transformations

Question

Recently I have been trying to understand extranatural and dinatural transformations. Here are the definitions I am working with.

From the nLab:

Let $F:A\times B\times B^{op}\to D$ and $G:A\times C\times C^{op}\to D$ be functors. A family of morphisms $$\{\alpha_{a,b,c}: F(a,b,b) \to G(a,c,c)\}_{a\in A,b\in B,c\in C} $$ is said to be ordinary natural in $a$ and extra-natural in $b,c$ if

1. Ordinary naturality in $a$: For all $b\in B, c\in C$ and morphisms $f:a\to a'$ in $A$, we have $\alpha_{a',b,c} \circ F(f,1,1) = G(f,1,1) \circ \alpha_{a,b,c}$
2. Extranaturality in $b$: For all $a\in A, c\in C$ and morphisms $g:b\to b'$ in $B$, we have $\alpha_{a,b,c} \circ F(1,1,g) = \alpha_{a,b',c}\circ F(1,g,1)$
3. Extranaturality in $c$: For all $a\in A, b\in B$ and morphisms $h:c\to c'$ in $C$, we have $G(1,h,1)\circ \alpha_{a,b,c} = G(1,1,h)\circ \alpha_{a,b,c'}$

Also from the nLab:

Let $F,G: C^{op}\times C \to D$ be functors. A dinatural transformation from $F$ to $G$ is a collection of morphisms $\alpha_c: F(c,c) \to G(c,c)$ such that for every morphism $f: c\to c'$ in $C$, we have $$G(1_c,f)\circ \alpha_c\circ F(f,1_c) = G(f,1_{c'}) \circ \alpha_{c'}\circ F(1_{c'},f)$$

Both of the above pages (and Wikipedia) claim that dinatural transformations generalise extranatural transformations, or in other words that extranatural transformations are a special case of dinatural transformations. However, I am struggling to see how this can be true. For example, for a dinatural transformation the domains of the functors $F$ and $G$ must be equal but for an extranatural transformation they do not. The most explicit description of the comparison given I can find is given here, where it is shown in Proposition 1.1.12 that an extranatural transformation $\alpha$ between $F,G: C\times C\times C^{op} \to D$ gives rise to a dinatural transformation between certain $F',G': A\times A^{op} \to D$ where $A:= C\times C^{op}\times C^{op}$. However, this only handles a special case for the domain of $F,G$.

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My questions are as follows:

1. Can the process described in Proposition 1.1.12 be extended to the case of $F: \mathcal{A}\times \mathcal{B}\times\mathcal{B}^{op}\to \mathcal{D}$, $G: \mathcal{A}\times \mathcal{C}\times\mathcal{C}^{op}\to \mathcal{D}$?
2. If so, how?
3. If not, why does one run into the unqualified statement that "extranatural is a special case of dinatural" so much? Is it because the special case of functors with domain $C\times C\times C^{op}$ is especially important?

Answer 1

$\newcommand{\A}{\mathcal{A}}\newcommand{\B}{\mathcal{B}}\newcommand{\C}{\mathcal{C}}\newcommand{\D}{\mathcal{D}}\newcommand{\E}{\mathcal{E}}\newcommand{\op}{{^{\mathsf{op}}}}$Fix $\A,\B,\C,\D$ and functors $F:\A\times\B\times\B\op\to\D,\,G:\A\times\C\times\C\op\to\D$.

Define $\E:=\A\times\B\times\C$. I will define two functors $F':\E\op\times\E\to\D,\,G':\E\op\times\E\to\D$ such that dinatural transformations $F'\to G'$ correspond to extranatural transformations $F\to G$.

$F'((a',b',c'),(a,b,c)):=F(a,b,b')$ and $G'((a',b',c'),(a,b,c)):=G(a,c,c')$ on objects; there is an obvious extension on arrows. It is a good and straightforward exercise to show my claim; there is a canonical bijection between dinatural $F'\to G'$ and extranatural $F\to G$.

Hint: It is actually very easy to actually define the corresponding extranatural transformation, given a dinatural transformation, and vice versa; if you're having trouble with that you're overthinking. Write down explicitly what the domain and codomain of these arrows should be in both cases. Then what remains to do is check these arrows actually are (extra, di)natural. To do so; I defined $F',G'$ essentially by taking some projections and then applying $F$ and $G$. Think about direct sums and components, you want to "pad your vector with zeroes", whatever that means, to prove the (di)(extra)naturality statements.