Are there adjoint functors that don't play nicely with internal homs?

Question

Let $\mathcal{C}$ be symmetric monoidal closed, with tensor product $- \otimes -$ and internal hom $[-,-]$.

In this case, we know that the tensor-hom adjunction internalizes, and $[X \otimes Y, Z] \cong [X, [Y,Z]]$ as objects in $\mathcal{C}$. Are there adjoint functors $L \dashv R$ from $\mathcal{C} \to \mathcal{C}$ for which this isn't true? That is, for which $[LX, Y] \not \cong [X, RY]$ in $\mathcal{C}$?

The obvious idea is to use yoneda:

$$ \begin{aligned} \mathcal{C}(A, [LX, Y]) &\cong \mathcal{C}(A \otimes LX, Y) \\ &\overset{\star}{\cong} \mathcal{C}(L(A \otimes X), Y) \\ &\cong \mathcal{C}(A \otimes X, RY) \\ &\cong \mathcal{C}(A, [X, RY]) \end{aligned} $$

But there's no reason a left adjoint should preserve tensor products, so I would expect step $\star$ to fail for many functors... Unfortunately, I'm struggling to come up with concrete examples where this fails.

Does anybody happen to know any? Obviously I would prefer "natural" examples (in the informal sense), preferably in $R$-mod or similar. Though I suspect the easiest examples will be found in heyting algebras viewed as poset categories.

Thanks in advance ^_^.

Answer 1

Suppose $\mathcal{C}$ is monoidal closed, with tensor product $- \otimes -$ and internal hom $[-,-]$. Suppose we have adjoint functors $L,R \colon \mathcal{C} \to \mathcal{C}$ that 'internalize', by which I mean there's a natural isomorphism

$$ [LX, Y] \cong [X, RY] .$$

Then up to natural isomorphism $L$ must be given by tensoring by some object $A$, and $R$ must be $[A , -]$. That is, there must exist natural isomorphisms

$$ LX \cong X \otimes A $$

and

$$ RX \cong [A, X]. $$

Conversely, any functor of the form $LX \cong X \otimes A$ does internalize in this way.

To see these facts, first remember the proof that any functor of the form $LX \cong X \otimes A$ does internalize:

$$ [X \otimes A, Y] \cong [X, [A, Y]]. $$

By the Yoneda lemma, this is equivalent to

$$ \mathrm{hom}(Z, [X \otimes A, Y]) \cong \mathrm{hom}(Z, [X, [A, Y]]) .$$

where the isomorphism is natural in all four arguments. Using the hom-tensor adjunction once on each side, this is equivalent to

$$ \mathrm{hom}(Z \otimes X \otimes A, Y) \cong \mathrm{hom}(Z \otimes X, [A, Y]]) $$

and using it again on the right side, this is equivalent to

$$ \mathrm{hom}(Z \otimes X \otimes A, Y) \cong \mathrm{hom}(Z \otimes X \otimes A, Y) $$

so we're done!

Now let's try to copy this argument starting from an arbitrary pair of adjoint functors $L, R \colon \mathcal{C} \to \mathcal{C}$. When do we have a natural isomorphism

$$ [LX, Y] \cong [X, RY] ? $$

By the Yoneda lemma, this is equivalent to

$$ \mathrm{hom}(Z, [LX, Y]) \cong \mathrm{hom}(Z, [X, RY]) .$$

where the isomorphism is natural in all four arguments. Using the hom-tensor adjunction once on each side, this is equivalent to

$$ \mathrm{hom}(Z \otimes LX, Y) \cong \mathrm{hom}(Z \otimes X, RY) $$

and using the adjunction between $L$ and $R$, this is equivalent to

$$ \mathrm{hom}(Z \otimes LX, Y) \cong \mathrm{hom}(L(Z \otimes X), Y) $$

Using the Yoneda lemma again, this is equivalent to

$$ Z \otimes LX \cong L(Z \otimes X) \qquad (\star) $$

In short, the existence of a natural isomorphism $[LX, Y] \cong [X, RY]$ is equivalent to $(\star)$.

But if $(\star)$ holds, we can take $X$ to be the unit of the tensor product and see

$$ Z \otimes LI \cong L(Z \otimes I) \cong LZ $$

It follows that up to natural isomorphism, $L$ must be given by tensoring with the object $A = LI$:

$$ LZ \cong Z \otimes A$$

Note that we never needed $\mathcal{C}$ to be symmetric.

Answer 2

Following John's answer, which gives a criterion for this internalization to work out, it's easy to answer the other half of this question:

Does anybody happen to know any [counterexamples]? Obviously I would prefer "natural" examples (in the informal sense), preferably in 𝑅-mod or similar. Though I suspect the easiest examples will be found in heyting algebras viewed as poset categories.

According to John's answer, it suffices to find a left adjoint which is not tensoring with something.

If $H$ is any heyting algebra (we can take $H = \{0,1\}$ for concreteness) then $A = H^\mathbb{N}$ is also a heyting algebra. Thus $A$ is cartesian closed as a (thin) category. Since $A$ has products (read: meets), we have an adjunction $(\Delta : A \to A^2) \dashv (- \times - : A^2 \to A)$. But $A^2 \cong A$ by the usual "shuffle" map

$$ \big((a_0,a_1,a_2,\ldots),(b_0,b_1,b_2,\ldots) \big) \mapsto (a_0,b_0,a_1,b_1,a_2,b_2,\ldots) $$

So composing, we see that we have an adjunction $\Delta \dashv \times$ where

$$ \Delta(a_0, a_1, a_2, \ldots) = (a_0, a_0, a_1, a_1, a_2, a_2, \ldots) $$

$$ \times(a_0, a_1, a_2, \ldots) = (a_0 \land a_1, a_2 \land a_3, \ldots) $$

From here it's easy to see that the left adjoint $\Delta$ cannot be of the form $\vec{a} \times -$ for any $\vec{a} \in A$. After all, $\Delta(1,1,1,\ldots) = (1,1,1,\ldots)$ would force $\vec{a} = (1,1,1,1,\ldots)$, but $(1,1,1,1,\ldots) \times -$ is the identity functor, while $\Delta$ isn't.

Since $\Delta$ is a left adjoint that is not of the form $\vec{a} \times -$, John's answer shows that it must be a counterexample! Indeed let $\vec{a} = (0,1,\ldots)$ and $\vec{b} = (1,1,0,0,\ldots)$. Then

$$ \begin{align} (\Delta \vec{a}) \Rightarrow \vec{b} &= (0,0,1,1,\ldots) \Rightarrow (1,1,0,0,\ldots) \\ &= (0 \Rightarrow 1, 0 \Rightarrow 1, 1 \Rightarrow 0, 1 \Rightarrow 0, \ldots) \\ &= (1, 1, 0, 0, \ldots) \end{align} $$

but

$$ \begin{align} \vec{a} \Rightarrow (\times \vec{b}) &= (0,1,\ldots) \Rightarrow (1,0,\ldots) \\ &= (0 \Rightarrow 1, 1 \Rightarrow 0, \ldots) \\ &= (1, 0, \ldots) \end{align} $$

so $(\Delta \vec{a}) \Rightarrow \vec{b} \ \neq \ \vec{a} \Rightarrow (\times \vec{b})$, and indeed this adjunction does not internalize!