I sometimes need to prove a category is Tannakian. Part of the definition of a Tannakian category is that it is rigid. 


However, I find the [definition](http://en.wikipedia.org/wiki/Rigid_category) of rigid categories somewhat difficult. I don't know how to show that these morphisms are identities.


Is there a way of looking at this that makes it more clear when these identities hold?


In my cases it is obvious that there exists a functor $D$ and isomorphisms:


$$Hom(X,Y) = Hom(D(Y),D(X))$$


$$Hom(X,Y)=Hom(1, Y \otimes D(X) ) $$


$$ X = D(D(X)) $$


$$ D(X \otimes Y) = D(X) \otimes D(Y)$$


So in particular I get maps $1 \to X \otimes D(X)$ and $X \otimes D(X) \to 1$. Is there a simple way of expressing duality in terms of a property of this functor?



Establishing Duality in Tannakian Categories
As Mostafa points in the comments, it suffices to have a canonical isomorphism
$$ \hom(X\otimes Y,Z) \overset?= \hom(X,Z\otimes D(Y)). $$


But if I am not mistaken, you have
$$ \begin{aligned} \hom(X\otimes Y,Z) & = \hom(1,D(X\otimes Y)\otimes Z) \\ & = \hom(1,D(X)\otimes D(Y)\otimes Z) \\ & = \hom(X, D(Y) \otimes Z)\end{aligned}$$
using what you know (and symmetric monoidality, which I assume you have, as otherwise I would have expected $D(X\otimes Y) = D(Y) \otimes D(X)$).


Note that I only used your second and fourth properties. I'm pretty sure that the second property alone is not good enough: you need some way to deal with tensor products in the first slot of $\hom$.


