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If a category $\mathcal{C}$ is (symmetric) monoidal closed, is the opposite category $\mathcal{C}^{\text{op}}$ also monoidal closed?

It is not clear to me whether by dualising the natural bijection $$\mathcal{C}(a\otimes b,c)\cong\mathcal{C}(a,[b,c])$$ we get a monoidal closed structure on $\mathcal{C}^{\text{op}}$.

PeteBazz
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1 Answers1

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From either definition of adjunction it follows that $F ⊣ G$ iff $G^{\mathrm{op}} ⊣ F^{\mathrm{op}}$. For a monoidal category to be monoidal closed it is enough that every $- ⊗ B$ be left adjoint. The dual of this is that $- ⊗ B$ is right adjoint. So no, the dual of a monoidal closed category is not itself closed (instead it is "coclosed", although this is not an often used term).

In fact, the dual of a Cartesian closed category is never closed, unless the category is the singleton: see here for a short proof.

user54748
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    Thanks! Regarding your last sentence, is it however possible that the dual of a monoidal closed category is monoidal closed? If yes, are you aware of any conditions on $\mathcal{C}$ that force this to happen? – PeteBazz Jul 04 '16 at 06:31
  • @PeteBazz Well it happens for the terminal category, so it's not impossible, although it certainly doesn't happen often. If the dual of $\mathcal C$ is closed, then $\mathcal C$ is coclosed, ie. tensoring is right adjoint, and in particular it preserves limits. So you'd need to find a monoidal category for which tensoring preserves both limits and colimits, and I don't know of any example. – user54748 Jul 04 '16 at 08:38
  • This answer however is interesting; it describes a coclosed subcategory of a closed category, but I don't know if this subcategory is itself still closed. – user54748 Jul 04 '16 at 08:40