Natural ways to make a functor adjoint

Question

Let $F: C \to D$ be a functor between two categories without a right adjoint. What are some natural ways to create a right adjoint for $F$?

Of course, this does not make sense on the nose. One needs to replace the categories by some other categories that are close enough. So the real question is:

Let $F: C \to D$ be a functor between two categories. Can one find another functor $\hat{F}: \hat{C} \to \hat{D}$ such that $\hat{F}$ admits a right-adjoint $\hat{G}$, such that

1. Solely from the data of $\hat{F}$, one can recover the data of $F$. Namely, there is a construction $\tilde{(\cdot)}$ such that $F \sim \tilde{\hat{F}}$, $C \sim \tilde{\hat{C}}$, and $D \sim \tilde{\hat{D}}$.
2. If $F$ admits a right-adjoint $G$ already, then $\tilde{\hat{G}} \sim G$.

It would be even better if $\hat{(\cdot)}$ is a localization of categories.

Answer 1

This is a bit too long to be a comment, but hopefully it can shed light on your question.

Adjointness via representability

The concept of adjointness, like many others in category theory, can be understood in terms of representable functors.

For any functor $\mathcal{C} \xrightarrow{F} \mathcal{D}$, we can define a functor \begin{alignat}{2} \mathcal{D} &\xrightarrow{G^{\mathsf{formal}}} &&[\mathcal{C}^{\mathop{op}}, \mathbf{Set}] \\ d &\longmapsto &&(c \mapsto \mathcal{D}(F c, d)). \end{alignat}

This looks a bit like the Yoneda embedding of $\mathcal{D}$, except its target is the presheaf on $\mathcal{C}$ which first applies $F$ and then takes the Hom functor in $\mathcal{D}$ fixed in its second argument.

Now we can claim the following.

Proposition. $F$ has a right adjoint if and only if $G^{\mathsf{formal}} (d)$ is representable for every object $d$ of $\mathcal{D}$.

For the forwards direction, assuming that $F \dashv G$, we get the representability of any functor in the image of $G^{\mathsf{formal}}$ by the natural bijection $\mathcal{D} (F c, d) \cong \mathcal{C} (c, G d)$. For the reverse direction, define the right adjoint to $F$ to be $Y^{-1} \circ G^{\mathsf{formal}}$ where $Y^{-1}$ is the inverse to the Yoneda embedding defined only on the representable functors $[\mathcal{C}^{\mathop{op}}, \mathbf{Set}]$ (every such one is in the essential image of $Y$). Then deduce $$ \mathcal{C}(c, Y^{-1} G^{\mathsf{formal}} d) \cong [\mathcal{C}^{\mathop{op}}, \mathbf{Set}](Y c, G^{\mathsf{formal}} d) \cong G^{\mathsf{formal}} d (c) = \mathcal{D} (F c, d), $$ naturally in $c$ and $d$, where the first step uses the adjoint equivalence $Y \dashv Y^{-1}$ and the second step is the Yoneda lemma.

This means that the right adjoint $G$ is a universal solution to the equation $\mathcal{D} \xrightarrow{G} \mathcal{C} \xrightarrow{Y} [\mathcal{C}^{\mathop{op}}, \mathbf{Set}] \cong \mathcal{D} \xrightarrow{G^{\mathsf{formal}}} [\mathcal{C}^{\mathop{op}}, \mathbf{Set}]$. The formal right adjoint to $F$, $G^{\mathsf{formal}}$, always exists, and it tells you how to build the $G$ under certain conditions.

A tangent on profunctors

If you like, you can see $G^{\mathsf{formal}}$ as a profunctor by uncurrying, and this leads to the (slightly disingenuous) slogan:

Every functor has a right adjoint, but it is a profunctor instead of a functor.

Now, up to Cauchy-completeness, every profunctor which admits a right adjoint in $\mathbf{Prof}$ is equivalent to a functor; analogously, the profunctor obtained from $G^{\mathsf{formal}}$ is equivalent to some functor $G$ exactly when it is a right-adjoint profunctor. This leads to an enhanced slogan:

Every functor has a right adjoint, but sometimes it is profunctor instead of a functor.

In this case, 'sometimes' means that it does not have a right adjoint in the usual sense.

Answer 2

I'll address the dual question, about functors without a left adjoint.

Let $G: D \to C$ be a functor that doesn't necessarily have a left adjoint. Under fairly mild conditions, $G$ has a codensity monad $T^G$. This is a monad on $C$, and it's what you think it is in the case where $G$ does have a left adjoint. The slick way to define $T^G$ is as the right Kan extension of $G$ along itself.

The forgetful functor $U^G: C^{T^G} \to C$ (where the domain is the category of $T^G$-algebras) is the universal monadic functor generated by $G$. A bit more precisely, there is a canonical comparison functor $D \to C^{T^G}$, and this exhibits $U^G: C^{T^G} \to C$ as the initial monadic functor into $C$ equipped with a map from $G: D \to C$ in the slice category $\mathbf{CAT}/C$. Draw a triangle to see what I mean!

What this construction does is replace your original functor $G$ by a functor $U^G$ that not only has a left adjoint but is also monadic, in a universal way. So that's more than you asked for. But perhaps it's relevant to your situation.

Of course, there's a dual construction for right adjoints, involving density comonads.

A reference: my paper Codensity and the ultrafilter monad, Theory and Applications of Categories 28 (2013), no. 13, 332-370.