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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 $-\otimes DX$?

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  • $\begingroup$ 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: $Map(Y,Z\otimes X)\cong Map(Y,Z\otimes DDX)$ ? $\endgroup$ – frabala Feb 14 '14 at 13:14
  • $\begingroup$ $X\cong DDX$ if $X$ is dualizable. A proof can be found in many articles/books introducing the notion of a dualizable object. $\endgroup$ – user8463524 Feb 14 '14 at 13:24
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    $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Feb 14 '14 at 14:30
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[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).

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

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