# Rigid monoidal and closed monoidal categories

I am trying to understand the relationship between rigid monoidal categories and closed monoidal categories. First every rigid monoidal category is closed, with an adjoint to the functor $$X \otimes -$$ given by $$X^* \otimes -$$.

Let $$\mathcal{C}$$ be a closed monoidal category (i.e., with internal homs), such that for all $$X \in \mathcal{C}$$, the functor $$X \otimes -$$ and its adjoint forms an equivalence of the category $$\mathcal{C}$$ with itself. Does it follow that $$\mathcal{C}$$ is rigid?

• I don't know the answer, but does Warning 1.2 at this link help? ncatlab.org/nlab/show/dual+object+in+a+closed+category – David White Aug 3 '20 at 13:46
• I don't understand what your question is. What does "adjoints of all $X\in\mathcal{C}$ form an equivalence" mean? The right adjoint to $X\otimes -$ is a functor $\mathcal{C}\to \mathcal{C}$, are you asking whether these are equivalences? In that case the answer is no... – Denis Nardin Aug 5 '20 at 11:32
• I think he means: "suppose $C$ is closed monoidal and for every $X$, the adjoint of $X\otimes -$ is an equivalence. Does it follow that $C$ is rigid?" – David White Aug 5 '20 at 11:43
• Yes this is what I mean. I have reworded to make things clearer. – Jake Wetlock Aug 5 '20 at 11:44
• Obviously, given such a closed monoidal category $C$, and given $X$, a candidate for $X^*$ is $Hom(X,1)$. But as I commented before, it's not clear why there should be a coevaluation morphism $1 \to X\otimes X^*$ in general. How does $Hom(X,-)$ being an equivalence help? – David White Aug 5 '20 at 11:50

Let $$1$$ be the unit of $$C$$. For every $$X$$, we define $$X^* = Hom(X,1)$$. I will assume $$C$$ is strict closed symmetric monoidal. Further assuming the condition the OP specified, we can show $$C$$ is rigid.

Let's unpack the additional condition the OP wants to assume. For every $$X$$, the functors $$F(-) = X\otimes -$$ and $$G(-) = Hom(X,-)$$ form an adjoint equivalence. In particular, the counit of the adjunction, $$\epsilon: FG \to 1_C$$ is a natural isomorphism. So, $$\epsilon_1: X\otimes X^* = X\otimes Hom(X,1) \to 1$$ is an isomorphism. Call this morphism $$ev_X$$. Define the coevaluation as its inverse.

Following Section 2.10 of Tensor Categories, we must show that the compositions:

$$X \stackrel{coev_X\otimes id}{\longrightarrow} (X\otimes X^*) \otimes X \stackrel{\alpha}{\longrightarrow} X\otimes (X^*\otimes X) \stackrel{id\otimes ev_X}{\longrightarrow} X$$, and

$$X^*\stackrel{id \otimes coev_X}{\longrightarrow} X^*\otimes (X \otimes X^*) \stackrel{\alpha^{-1}}{\longrightarrow} (X^*\otimes X) \otimes X^* \stackrel{ev_X\otimes id}{\longrightarrow} X^*$$

are the identity morphisms. But the first is just $$X\cong 1\otimes X$$, followed by the associator, followed by $$X\otimes 1 \cong X$$, which is certainly the identity on $$X$$ (by strictness), and the second works the same way. So, indeed, $$X^*$$ is a left dual to $$X$$.

By symmetry, $$X^*$$ is also the right dual to $$X$$, and $$X^*\otimes X \cong 1$$, so the two morphisms in 2.10.2 of Tensor Categories are also identities. Hence, every $$X$$ has both a left and right dual, so $$C$$ is rigid.

• Stupid question: Is an adjoint equivalence the same as an adjunction between equivalences? – Bipolar Minds Aug 5 '20 at 13:41
• I just meant that in the usual adjunction you get from being closed monoidal, the unit and counit are natural isomorphisms (like the OP wanted), so that the adjunction is in fact an equivalence. Hopefully the term "adjoint equivalence" doesn't have alternative meanings in the tensor category world. ncatlab.org/nlab/show/adjoint+equivalence – David White Aug 5 '20 at 13:45
• As it says here, every equivalence can be "improved" to an adjoint equivalence, and it just comes down to which morphisms you pick to be the unit and counit. So it is no loss of generality. math.stackexchange.com/questions/595482/… – David White Aug 5 '20 at 13:47