Localization of cocomplete categories and right orthogonality: does the equivalence always hold?

107 Views Asked by At

In Handbook of Categorical Algebra, Volume 1: Basic category theory, Borceux proves the following (around theorem 5.4.7 page 198).

Definition. An object $x$ of a category $\newcommand{\cC}{\mathsf{C}}\cC$ is called $α$-compact for some (regular) cardinal $α$ if $\newcommand{\Hom}{\mathrm{Hom}}\Hom(x,—)$ preserves $α$-filtered colimits.

Definition. An object $x$ of a category $\cC$ is right-orthogonal to a morphism $f : a→b$ if $\Hom(f,x) : \Hom(b,x) → \Hom(a,x)$ is a bijection. ($x$ sees $f$ as an isomorphism from the right.)

Theorem. Let $\cC$ be a cocomplete category and let $Σ$ be a class of morphisms in $\cC$ whose every domain and codomain is $α$-compact for some $α$. Then the category of objects right-orthogonal to every morphism in $Σ$ is reflective and the reflector is the cocontinuous localization of $\cC$ at $Σ$.

The cocontinuous localization means that it is the universal cocomplete category $\newcommand{\cD}{\mathsf{D}}\cD$ equipped with a functor $\cC → \cD$ sending every morphism in $Σ$ to an isomorphism.

I was wondering if we also have the following (or counter-examples).

Question. Let $\cC$ be a cocomplete category and let $Σ$ be a class of morphisms in $\cC$. Then the following are equivalent for a reflective full subcategory $\cD$ :
(1) The reflector $\cC → \cD$ is the cocontinuous localization of $\cC$ at $Σ$.
(2) $\cD$ is the set of objects right orthogonal to every morphism in $Σ$.

The techniques used in the proof of Borceux rely heavily on the construction of the reflector (which itself is an adaptation of the similar construction for posets instead of categories). I think they cannot be applied here.


If every cocontinuous morphism out of $\cC$ has a right adjoint, then I know how to prove that (2) $⇒$ (1). Also, (1) implies that every element of $\cD$ is right orthogonal to evert arrow in $Σ$, but the question is to know if the converse holds (I don't know the answer even if we suppose the existence of right adjoints).

Here is a proof that (2) $⇒$ (1) if $\cC$ satisfies the conditions of the special adjoint functor theorem.

Proof. Suppose (2). Let $\newcommand{\cE}{\mathsf{E}}\cE$ be a cocomplete category and let $F : \cC → \cE$ be a cocontinuous functor such that $F(f)$ is invertible for all $f ∈ Σ$. Then the right adjoint $G : \cE→\cC$ of $F$ takes values in $\cD$, because for all $f ∈ Σ$, we have $\Hom(f,G(x)) = \Hom(F(f),x)$ and $F(f)$ is an iso. Let $M$ be the monad induced by $\cD ⊆ \cC$. Then $G$ has the structure of a right $M$-algebra and then its left adjoint $F$ has the structure of a left $M$-algebra, which means that $F$ factorizes through $\cC → \cD$.

I tried to transfer this reasoning to the case when $F$ only has a "formal adjoint" which is a profunctor, but the problem is that I don't know if this formal adjoint is automatically a right $M$-algebra in this setting... It satisfies a property similar to "taking values in $\cD$" but I don't believe it is enough to have a right $M$-algebra.

I don't have examples of this situation so I don't have a "real reason" to ask this question, but I still would like to know.