# Contravariant internal hom

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.

• 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 '18 at 11:00
• Sorry! I really meant $[-,A]$ preserves the empty set (limit in $\textbf{Set}^{op}$) iff $A=\varnothing$. – Cat John Nov 9 '18 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.
• 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 '18 at 10:13