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 $-\otimes DX$?

  • $\begingroup$ 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
  • 5
    $\begingroup$ 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

[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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.