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

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.

• For module categories, every pair of adjunction is naturally isomorphic to a tensor-hom adjunction with fixed module. But Heyting algebras might be a good idea indeed. Oct 1, 2021 at 7:27
• @Berci -- that's really surprising! It makes me feel better about my inability come up with other examples though, haha. Do you happen to have a reference? Oct 1, 2021 at 7:33
• @HallaSurvivor Berci is referring to the Eilenberg-Watts theorem. As for your step $\star$, it seems that what you need is an isomorphism $A\otimes L(X)\simeq L(A\otimes X)$, which makes me think of tensorial strength, but I don't know how helpful it is. Oct 1, 2021 at 7:50
• The property – or rather structure – of preserving tensor products in the sense you are interested in is related to tensorial strength. It is indeed not automatic. You can check that the subdivision functor $\textrm{Sd} : \textbf{sSet} \to \textbf{sSet}$ is a left adjoint that does not have a tensorial strength that is an isomorphism. Oct 1, 2021 at 13:15
• Probably worth adding that the isomorphism is meant to be natural, or it would be quite easy to construct a counterexample. Feb 14 at 23:10

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.

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!