This is from Lemma 3.7 in $SL(2,\mathbb{Z})$-action for ribbon quasi-Hopf algebras.

Let $\mathcal{C}$ be a braided monoidal category with left duals (small, strict), define the functor $F=-^*\otimes-$, and let $\mathcal{D,N}:\mathcal{C}\to\mathbf{Set}$ be the functors which send objects $V\in\mathcal{C}$ to

  • $\mathcal{D}V = \mathrm{Din}(F,V)$, the set of dinatural transformations $j:F\xrightarrow{..}V$
  • $\mathcal{N}V=\mathrm{Nat}(1,1\otimes V)$, the set of natural transformations $\eta:1\Rightarrow1\otimes V$

For objects $V\in\mathcal{C}$, define maps $\zeta_V:\mathcal{D}V\to\mathcal{N}V$ by

$$ (\zeta_V(j))_X \equiv \left[ X=1X\xrightarrow{c_X\otimes 1}XX^*X\xrightarrow{1\otimes j_X} XV \right], $$ where $c_X$ is the coevaluation for $X$ and we omit the symbol $\otimes$ for readability reasons.

These maps are isomorphisms, which is easily seen, but my problem is the statement that

Naturality of $\zeta$ follows from dinaturality of $j$.

Let $f:V\to W$. We want to show naturality, that is $$ \mathcal{N}f\circ \zeta_V = \zeta_W \circ \mathcal{D}f, $$ where the functors act on morphisms in the obvious way. It is sufficient to check this on elements and components, so we compute \begin{align*} (\mathcal{N}f\circ \zeta_V)(j)_X &= (1\otimes f)\circ(\zeta_V(j))_X \\ &= (1\otimes f)\circ (1\otimes j_X)\circ (c_X\otimes 1) \\ &= (1\otimes (f\circ j_X))\circ (c_X\otimes 1)\\ &= (1\otimes \mathcal{D}f(j)_X)\circ (c_X\otimes 1)\\ &= (\zeta_W \circ \mathcal{D}f)(j)_X\ , \end{align*} which is exactly what we wanted to show.

But I didn't use the dinaturality of $j$ (at least I'm not aware of). Where's my mistake?


I want to make clear what dinaturality means:

$j:F\xrightarrow{..} V$ iff for all $A\in\mathcal{C}$ we have morphisms $j_A:A^*\otimes A\rightarrow V$ s.t. for all $f:A\to B$ the diagram $$\require{AMScd} \begin{CD} B^*\otimes A @>{f^*\otimes 1}>>A^*\otimes A \\ @V{1\otimes f}VV @V{j_A}VV \\ B^*\otimes B @>{j_B}>> V \end{CD} $$ commutes.

I don't see where this comes in.


I agree that this seems like a mistake. It looks clear to me that $\zeta$ can be extended to a natural isomorphism between the sets of not-necessarily dinatural transformations $F\to V$ and not-necessarily natural transformations $1\to 1\otimes V$; after all, this is essentially just the natural isomorphism $[X^*X,V]\cong [X,XV]$, with a product taken over all $X$, no? The point should be that naturality of $\zeta_V(j)$ follows from dinaturality of $j$.

  • $\begingroup$ Thank you. Without thinking too much, I can appreciate "after all, this is essentially just the natural isomorphism $[X^*X,V]\cong [X,XV]$, with a product taken over all $X$". And indeed, the naturality from $\zeta_V(j)$ then follows from the dinaturality of both $j$ and the coevaluation (the dinaturality of the latter is also known as sliding to the right). But, besides that, the computation is not enlightening, so I won't carry it out here. $\endgroup$ – Jo Be Apr 10 '18 at 8:06
  • $\begingroup$ Actually, using string diagrams this is a really fast calculation, but I can't draw them here. $\endgroup$ – Jo Be Apr 10 '18 at 8:26

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