Unitors in star-autonomous categories 1.Context
Let $(C, \otimes,  I, a, l,r)$ be a monoidal category.
Suppose $S: C^{op} \xrightarrow{\sim} C$ is an equivalence of categories with inverse $S’$. Assume that there are bijections $\phi_{X,Y,Z}: Hom_C(X \otimes Y,SZ) \xrightarrow{\sim} Hom_C(X, S(Y \otimes Z))$ natural in $X,Y,Z$. (This makes $C$ a star-autonomous category.) Note that we do not assume that $S$ is a monoidal equivalence. For simplicity suppose that the associator $a$ is the identity and that $S$ and $S'$ are strict inverses.
2.Question

Is the morphism $$\phi_{B,I,S'B}^{-1}\big(S(l_{S'B})\big):B \otimes I \rightarrow SS'(B)=B$$ equal to the right unitor $r_B$ for every $B \in C$?

Some ideas and follow-up questions:

*

*This equality would imply that in star-autonomous categories a choice of one unitor immediately determines the other.


*Does $\phi$ map isomorphisms to isomorphisms?
At least for $\phi^{-1}$ I know this not to be true: Note that a star-autonomous category is monoidal closed with left internal hom $[X,Y]:=S(X \otimes S'Y)$. Even though the map $id_{S(X \otimes S'Y)}=id_{[X,Y]}$ is invertible, the morphism $\phi^{-1}_{id_{[X,Y]},X,S'Y} (id_{[X,Y]})$ is the evaluation morphism $ev_{X,Y}$ which is in general not an isomorphism.
Maybe one can show that $\phi_{B,I,S'B}(r_B): B \rightarrow S(I \otimes S'B)$ is not in general an isomorphism. This would give a negative answer to the above question. The Yoneda lemma (covariant version) tells us something about the form of natural transformations between functors $F,Hom(-,X): C^{op} \rightarrow Set$. Can it be modified to cover natural transformations between functors $C^{op} \times C^{op} \times C^{op} \rightarrow Set$ as above?


*Does there exist a property (for example the requirement that the unitor makes a certain diagram commute) that characterizes the left/right unitor uniquely? This would indicate a strategy for giving a positive answer: One could try to verify that $\phi_{B,I,S'B}^{-1}\big(S(l_{S'B})\big)$ satisfies the characteristic property.
Such a property could only hold in special cases (for instance for star-autonomous categories) since in general monoidal categories unitors are a chosen structure not a property.


*What is a good place to look for counterexamples to the equality?


*

*I tried rigid monoidal categories – to no avail:
Consider for instance the category $vect_{\mathbb F}$ of finite-dimensional vector spaces over a field $\mathbb F$ with usual tensor product. Let $S=S':=(-)^*$ be the duality functor. Define $\phi$ on $X, Y, Z \in vect_{\mathbb F}$ as $$\Big( \big(\phi_{X,Y,Z}(k)\big)(x) \Big)(y\otimes z):=\big(k(x\otimes y)\big)(z)$$ for $k:X \otimes Y \rightarrow Z^*$. Then $\phi$ is natural in all three components and invertible for any $X,Y,Z \in vect_{\mathbb F}$ for reasons of dimension. Denote by $\iota_X: X \rightarrow X^{**}$ the canonical identification of $X \in vect_{\mathbb F}$ with its double dual. One shows that $l_{X^*}^* \circ \iota_x=\phi_{X,\mathbb F, X^*}(\iota_x \circ r_x)$ for all $X \in vect_{\mathbb F}$ by evaluating both sides first on an element $x \in X$ and then on simple tensors in $\mathbb F \otimes X^*$.

*Similarly, the  category of quadratic algebras presented on the nLab satisfies the above equation. This is because the natural transformation $\phi$ is essentially the one for finite-dimensional vector spaces.

*The semicartesian *-autonomous category related to Łukasiewicz logic satisfies the above equation since it is a posetal category.



*

*A hopefully useful observation:
Using the naturality of $r$ and $\phi$ I was able to show that the above statement is equivalent to the claim that the following equality
$$\phi_{S(I\otimes S'B), I,S'B}\big( S(l_{S'B}^{-1})\circ r_{S(I \otimes S'B)}\big)=id_{S(I\otimes S'B)}$$ holds. Maybe this observation is of help.

 A: Denote by $\eta: 1_C \Rightarrow SS'$ the natural isomorphism that is part of the data of the equivalence of categories $S: C^{op} \xrightarrow{\sim} C$. It is in general not true that the equality $$S(l_{S'X})\circ \eta_X=\phi_{X,I,S'X}\big(\eta_X \circ r_X\big)$$ holds for every $X \in C$. I will construct a counterexample by modifying the standard $\ast$-autonomous structure on $FinVect_{\mathbb K}$.
FinVect is $\ast$-autonomous
Consider the category $C:=FinVect_{\mathbb K}$of finite dimensional vector spaces over a field $\mathbb K$. Let $S=S'$ be the standard (contravariant) duality functor, i.e. $S(X):=X^*$ for $X \in C$ and $S(f):=f^*$ where $f^*$ denotes the transpose of the linear map $f$. Since we only consider finite-dimensional vector spaces the functor $S$ is an antiequivalence with inverse $S$.
Let $X,Y,Z \in C$. For every $x \in X$ and $f \in C(X \otimes Y,Z^*)$ define $\phi_{X,Y,Z}(f)(x)\in (Y \otimes Z)^*$ as the unique linear map that corresponds under the universal property of the tensor product to the bilinear map $$Y \times Z \rightarrow \mathbb K \\ (y,z) \mapsto f(x \otimes y)(z).$$ One shows that for every $f \in C(X \otimes Y,Z^*)$ the following map is linear $$\phi_{X,Y,Z}(f): X \rightarrow (Y \otimes Z)^* \\ x\mapsto \phi_{X,Y,Z}(f)(x).$$
Set $$\phi_{X,Y,Z}: C(X \otimes Y,Z^*) \rightarrow C(X,(Y \otimes Z)^*) \\ f \mapsto \phi_{X,Y,Z}(f).$$
One shows:

*

*The function $\phi_{X,Y,Z}$ is linear and has trivial kernel. By rank-nullity this proves that $\phi_{X,Y,Z}$ is a bijection.

*The function $\phi_{X,Y,Z}$ is natural in all components. Unpacking definitions this precisely means that $$(g\otimes h)^* \circ \phi_{X,Y,Z}(k) \circ f= \phi_{X,Y,Z}(h^* \circ k \circ(f \otimes g))$$ for all $f:E \rightarrow X, g:F \rightarrow Y,h:G \rightarrow Z, k:X\otimes Y \rightarrow Z^*$.

This makes $(C,S,S', \phi)$ a $\ast$-autonomous category. One verifies that with $l_X(\lambda \otimes x)= r_X(x \otimes \lambda)=\lambda \cdot x$ the equality in question holds for this $\ast$-autonomous category.
Omitting the associator we equivalently could have defined $$\phi_{X,Y,Z}(f):X \xrightarrow {r_X^{-1}}X \otimes \mathbb K \xrightarrow{\text{id}_X \otimes \text{coev}_Y} X \otimes Y \otimes Y^* \xrightarrow{f\otimes \text{id}_{Y^*}}Z^*\otimes Y^* \xrightarrow {\sim} (Y \otimes Z)^*$$ where $\text{coev}_Y: \mathbb K \rightarrow Y \otimes Y^*; \lambda \mapsto \lambda (\sum_{i=1}^n y_i\otimes y_i^*)$ is the usual coevaluation
with $\{y_1,\cdots,y_n\}$ a basis of $Y$ and $\{y_i^*,\cdots,y_n^*\}$ the corresponding dual basis. Note that the last definition works in any rigid monoidal category.
Counterexample: A different $\ast$-autonomous structure on FinVect
Now let $\mathbb K$ be a field with more than two elements. Let $\mu \in \mathbb K$ be invertible and different from $1_{\mathbb K}$.
Define $\widetilde{\phi_{X,Y,Z}}:=\mu \cdot \phi_{X,Y,Z}$.
Then $\widetilde{\phi_{X,Y,Z}}$ is a linear isomorphism with inverse $\mu ^{-1}\cdot \phi_{X,Y,Z}^{-1}$. Since $\phi_{X,Y,Z}$ is natural, so is $\widetilde{\phi_{X,Y,Z}}$. Therefore $(FinVect_{\mathbb K},S,S', \widetilde{\phi})$ is $\ast$-autonomous. However: $$ l_{X^*}^*\circ \eta_X = \phi_{X,I,X^*}\big(\eta_X \circ r_X\big)\neq \mu\cdot\big(\phi_{X,I,X^*}(\eta_X \circ r_X)\big)= \widetilde{\phi_{X,I,X^*}}\big(\eta_X \circ r_X\big).$$ Note that $\widetilde{\phi}$ is essentially $\phi$ but with each coevaluation scaled by the same invertible element $\mu$.
This answers the above question negatively. However, some questions remain:

*

*Can $\phi$ always be chosen such that the above equality holds?

*Can the above construction be modified to work for other $\mathbb K$-linear $\ast$-autonomous categories?

