# Characterisation of duals in rigid monoidal categories

In "On Fusion Categories" by Etingof, Nikshych and Ostrik Proposition 2.1 there is used the following characterisation of a right dual in a rigid monoidal category (let us restrict to strict monoidal categories for simplicity):

Let $V$ be a simple object in a rigid monoidal category. A simple object $Y$ is the left dual of $V$ if and only if there is an epimorphism $V \otimes Y \rightarrow 1$. Let us call this map $\phi$.

I know that left/right duals are unique, but the prove for $Y \simeq ^*X$ that I know, needs in addition (for the case of left duals) that there is a map $\psi: 1 \rightarrow Y \otimes X$ such that $X \stackrel{id \otimes \psi}{\rightarrow} X \otimes Y \otimes X \stackrel{\phi \otimes id}{\rightarrow} X$ is the identity.

Can somebody give me a hint or a source where the above uniqueness is proved? Thank you very much!

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Haven't you missed some further hypotheses like that the category is semi-simple $k$-linear and simplicity of both $V$ and $Y$? As you state it this seems horribly wrong. – t.b. Mar 16 '11 at 12:04
Actually in "Lectures on Tensor Categories" by Calaque and Etingof there is the statement "[But] in a rigid category, the only simple object $Y$ such that $V \otimes Y$ projects on $1$ is $^*V$. I forgot "simple object" and "projects" I guess. I will edit the question accordingly. – Sven Raum Mar 16 '11 at 12:28

A direct calculation shows that in a rigid tensor category $Hom(V \otimes Y, 1) \cong Hom(V, {}^*Y)$. (This is essentially the calculation that shows that the two definitions of adjoint functors agree.) In particular, if $V$ and $Y$ are simple, then Schur's Lemma says that $Hom(V, {}^*Y)$ is zero unless $V \cong {}^*Y$.