# Is dualizablility of an object equivalent to tensoring with that object having a left adjoint?

Let $C$ be a closed symmetric monidal category. There is hence an adjunction $$-\otimes X\colon C\leftrightarrows C\colon Map(X,-)$$ involving the internal Hom $Map(-,-)$ for every object $X$ of $C$

An object $X$ of $C$ is called dualizable if the canonical map $$X\otimes DX\to Map(X,X)$$ is an isomorphism where $DX=Map(X,1)$. It turns out, that this condition is equivalent to the condition that the canonical map $Y\otimes DX\to Map(X,Y)$ is an isomorphism for each $Y$ in $C$. The isomorphism $$Map(Y,Z\otimes X)\cong Map(Y,Z\otimes DDX)\cong Map(Y,Map(DX,Z))\cong Map(Y\otimes DX, Z)$$ shows that there is an adjunction $$-\otimes DX\colon C\leftrightarrows C\colon -\otimes X$$ for a dualizable $X$, so then $-\otimes X$ has not only a right adjoint but also a left adjoint.

Is an object $X$ of $C$ necessarily dualizable, if $-\otimes X$ has a left adjoint and does this left adjoint have to be $-\otimes DX$?

• I don't think I can contribute to an answer. But, your question seems interesting to me. There's just a part I haven't understood. Why is this: $Map(Y,Z\otimes X)\cong Map(Y,Z\otimes DDX)$ ? – frabala Feb 14 '14 at 13:14
• $X\cong DDX$ if $X$ is dualizable. A proof can be found in many articles/books introducing the notion of a dualizable object. – user8463524 Feb 14 '14 at 13:24
• Interesting question. If $C=\mathsf{Mod}(R)$, then the answer is yes, by the Eilenberg-Watts Theorem: The left adjoint of $X \otimes -$ is a cocontinuous functor, hence given by tensoring with some object - this has to be the dual of $X$. For general $C$ this argument doesn't work. – Martin Brandenburg Feb 14 '14 at 14:30

[This answer is copied with very minor modifications from a preprint of mine "The Balanced Tensor Product of Module Categories" joint with Chris Douglas and Chris Schommer-Pries which will appear on the arxiv in the next few weeks.]

Let $\mathcal{R} \cong \mathrm{Vec} \oplus \mathrm{Vec} \cdot X$ be the symmetric monoidal category consisting of pairs of vectors spaces which we write as $V_1+V_2 X$ with tensor product given by

$(V_1+V_2 X) \otimes (W_1+W_2 X) = (V_1 \otimes W_2 \oplus V_2 \otimes W_1)X.$

Up to equivalence there are unique choices of associator, unitors, and symmetric structure making this a symmetric monoidal category. It is both finite and semisimple, and is a categorification of the ring $k[x]/(x^2)$, but it is not rigid. The object X cannot have a dual as there is no object $Z \in \mathcal{R}$ such that $Z \otimes X$ has a non-zero map to or from the unit object of $\mathcal{R}$.

However, it is easy to see that the tensoring with $X$ is an exact functor and hence by the adjoint functor theorem has both adjoints. Explicitly, the adjoint to tensoring with $X$ is the functor which sends $X \mapsto 1$ and $1 \mapsto 0$ (since $\mathcal{R}$ is semisimple this describes a unique functor).

• This is a counterexample to all sorts of things. A fun exercise is to compute its Drinfeld center. – Noah Snyder Feb 27 '14 at 5:46