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?



Rigid monoidal and closed monoidal categories
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](http://www-math.mit.edu/%7Eetingof/egnobookfinal.pdf), 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.


