Natural isomorphism in linearly distributive categories with left and right dualities Let $\mathrm{C}$ be a linearly distributive category with a left and right duality, i.e. it is a monodical category "twice": once for the bifunctor $\otimes:\mathrm{C}\times\mathrm{C}\to\mathrm{C}$ with unit $e$, and again for the bifunctor $\bullet:\mathrm{C}\times\mathrm{C}\to\mathrm{C}$ with unit $u$ (both with natural associativity and identity morphisms and all distributivity laws), and with right duality information (an object $A$* and two morphisms $ax^R_A:e\to A^*\bullet A$ and $cut^R_A:A\otimes A^*\to u$ for every object $A$) and left duality information (an object *$A$ and two morphisms $ax^L_A:e\to A\bullet ^*A$ and $cut^R_A:^*A\otimes A\to u$ for every object $A$), (both comes again with coherence properties, for more information see https://www.irif.fr/~mellies/mpri/mpri-ens/biblio/categorical-semantics-of-linear-logic.pdf, pages 74-78, from where I'm trying to do the exercise).  
The problem is to prove the natural isomorphisms between $A, ^*(A^*),(^*A)^*$. The given hint suggests to show the natural in $A$ and $B$ bijections $\mathrm{C}(A,B)\cong \mathrm{C}(e,A^*\bullet B)\cong \mathrm{C}(^*(A^*),B)$.  
Take the first natural isomorphism equality. To show that, I assume, we consider two functors $\mathrm{C}(1st,2nd)$ and $\mathrm{C}(1st^*\bullet1st,1st^*\bullet2nd)$ and should show the natural isomorphism between them, and after we can probably relate $\mathrm{C}(e,A^*\bullet B)$ and $\mathrm{C}(A^*\bullet A,A^*\bullet B)$ through the $ax^R_A$ morphism. For simplicity we can do that in two steps: taking the argument $A$ to be fixed and then the same for the argument $B$. But I do not see how I should proceed to show this natural bijection.   
Will be glad for any help and suggestions.
 A: Use adjointness instead!
The assumption in the exercise is probably that you'll use the adjointness properties listed in propositions 4 and 5, in which case the result is almost immediate. However, if you want to do it explicitly, then you can.
Explicit bijection
I'm going to drop explicit mention of associativity, distributivity and unit isomorphisms, otherwise this will get really unwieldy.
You've constructed the correct map from $C(A,B)\to C(e,A^*\newcommand\blt\bullet\blt B)$, namely $f\mapsto (1_{A^*}\blt f)\circ \newcommand\ax{\operatorname{ax}}\ax^R_A$.
The inverse map is given $g:e\to A^*\blt B$, first send $g$ to 
$1_A\otimes g: A \to A\otimes A^*\blt B$, and then postcompose $\newcommand\cut{\operatorname{cut}}\cut^R_A\blt 1_B$ to get the map
$(\cut^R_A\blt 1_B) \circ (1_A\otimes g) : A\to B$.
Checking that these are inverse bijections involves the coherence properties.
For one direction, we have 
$$
\begin{align}
(\cut^R_A\blt 1_B)\circ &(1_A\otimes ((1_{A^*}\blt f)\circ \ax^R_A))\\
&=
(\cut^R_A\blt 1_B)\circ (1_A\otimes 1_{A^*}\blt f)\circ (1_A\otimes \ax^R_A)\\
&=f\circ (\cut^R_A\blt 1_A)\circ (1_A\otimes \ax^R_A)\\
&= f.
\end{align}
$$
The first equality is functoriality of $\otimes$.
The second is functoriality of $\blt$ twice,
$$(\cut^R_A\blt 1_B)\circ (1_A\otimes 1_{A^*} \blt f) = 
\cut^R_A\blt f = (1_u\blt f)\circ (\cut^R_A\blt 1_A),$$
and the last equality is one of the coherence properties.
The other composition order will be the same idea.
Again, use adjointness instead!
You should probably just use/prove the adjunctions instead, however, since this proof is just a special case of the proof that $A\otimes - $ is left adjoint to $A^* \blt -$.
(The proof of the adjunction in this case is essentially that $\ax^R_A\otimes -$ is the unit, and $\cut^R_A\blt -$ is the counit, and the triangle identities follow from the coherence properties, and all the other adjunctions are similar.)
