Let $\mathcal{C}$ a symmetric monoidal category. By using symmetry, it is very easy to show that the contravariant internal hom functor $[-,A]\colon\mathcal{C}^{op}\longrightarrow\mathcal{C}$ is adjoint on the right to itself.

My question is: when is this functor additionally $[-,A]$ a left-adjoint? Do such objects $A$ have a name? In the category of $\textbf{Set}$ it is easily seen that $[-,A]$ preserves the terminal object $\textbf{Set}^{op}$ iff $A$ is the empty set.

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  • 3
    Are you sure your last claim is correct? Unless I’m missing something, $[-,0]$ isn’t a left adjoint. If it were, then mapping out of the adjoint into 2 and noting that $0$ and $1$ have unique “square roots” in $\mathrm{Set}$ shows that the adjoint would be the functor $[-,0]$ itself. But $\mathrm{Set}([X,0],Y) \simeq \mathrm{Set}([Y,0],X)$ is not true in general: consider $X=0$, $Y = 2$. – Peter LeFanu Lumsdaine Nov 9 at 11:00
  • Sorry! I really meant $[-,A]$ preserves the empty set (limit in $\textbf{Set}^{op}$) iff $A=\varnothing$. – Cat John Nov 9 at 12:27

Suppose that everything is dualizable (as in the category of finite-dimensional vector spaces over a field, for example), and write $TX=[X,A]=X^*\otimes A$. Then \begin{align*} \mathcal{C}(TX,Y) &= \mathcal{C}(1,X\otimes Y\otimes A^*) = \mathcal{C}(Y^*\otimes A,X) = \mathcal{C}^{\text{op}}(X,TY) \\ \mathcal{C}(W,TX) &= \mathcal{C}(1,W^*\otimes X^*\otimes A) = \mathcal{C}(X,W^*\otimes A) = \mathcal{C}^{\text{op}}(TW,X) \end{align*} so $T$ is self-adjoint on both sides.

If $\mathcal{C}$ is Cartesian, then $A$ must be initial.

Indeed, let $G : \mathcal{C} \to \mathcal{C}^{\mathrm{op}}$ be the right adjoint to $[-,A]$. Then we have a natural bijection $\mathrm{Hom}_\mathcal{C}([X,A],Y) \simeq \mathrm{Hom}_{\mathcal{C}^{\mathrm{op}}}(X,G(Y))$. Let $X = 1$. Then $\mathrm{Hom}_\mathcal{C}([1,A],Y) \simeq \mathrm{Hom}_\mathcal{C}(A,Y)$. On the other hand, $\mathrm{Hom}_{\mathcal{C}^\mathrm{op}}(1,G(Y)) \simeq {*}$ since $1$ is initial in $\mathcal{C}^\mathrm{op}$. So, $A$ is initial.

As Peter noted in the comments, $[-,0]$ is not a left adjoint in $\mathrm{Set}$. To give an example of a Cartesian closed category in which this is true, this is true in any Boolean algebra considered as a category. In this case $[-,0]$ is simply the negation. So, $[[X,0],0] = X$ and $[-,0]$ is its own right adjoint.

  • 1
    Wait... $G(1)$ is terminal in $\mathcal{C}^{op}$, so it must be initial in $\mathcal{C}$ and you get no further condition on $G(1)$. Am I missing something? – Denis Nardin Nov 9 at 10:13
  • 2
    Oops. My mistake. I'll edit the answer – Valery Isaev Nov 9 at 10:16

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