This is [cross-posted from MSE](https://math.stackexchange.com/questions/3962953/can-the-category-of-s-local-objects-be-reflective-but-not-a-localization-by-s) (and substantially re-written) after receiving no answers.


Suppose $\mathcal C$ is a category and $S \subseteq \operatorname{Mor}(\mathcal C)$ is some collection of morphisms in $\mathcal C$. Let $\operatorname{Loc}(S) $ be the full subcategory of $\mathcal C$ on the $S$-local objects. Then we have
$$
\operatorname{Loc}(S) \stackrel{\iota}{\hookrightarrow} \mathcal C \stackrel{P}{\twoheadrightarrow} \mathcal C[S^{-1}],
$$
where $\iota$ is the inclusion and $P$ is the canonical functor to the localization of $\mathcal C$ by $S$. We can ask about two possibly-existing objects:


1. a left adjoint $L$ to $\iota$;
2. a right adjoint $r$ to $P$.


In fact, (2) "subsumes" (1), in the following sense: If such an $r$ exists, then there is an equivalence of categories $j : \mathcal C[S^{-1}] \to \operatorname{Loc}(S)$ such that $r \cong \iota \circ j$. This implies that $j \circ P \vdash \iota$, so we have our $L$. Suppressing $j$, we can summarize as "if $P$ has a right adjoint, it must be $\iota$" or "if (2) exists, it is $\iota$, and hence (1) does too, and is $P$".


It's a bit tempting (at least to me) to try to extract something symmetrical like "if (2) exists, it must be $\iota$, and if (1) exists, it must be $P$" from this last summary, or maybe even "(1) exists iff (2) does", but neither of these actually follows from the statements above—to show them, we'd need a separate result about (1) "subsuming" (2) in the manner above.


So I'd like to know: **Are the following equivalent statements true?**



> 
> * If $L \vdash \iota$, then there is an equivalence of categories $j : \mathcal C[S^{-1}] \to \operatorname{Loc}(S)$ such that $L \cong j \circ P$.
> * If $\iota$ has a left adjoint, then $P$ has a right adjoint.
> * If $L \vdash \iota$, then $L$ exhibits $\operatorname{Loc}(S)$ as a localization of $\mathcal C$ by $S$.
> 
> 
> 


I suspect the answer is "not necessarily", because [the nLab page on reflective localizations](https://ncatlab.org/nlab/show/reflective+localization) lists two facts which would clearly be implied by the third statement above but which do not clearly imply it. In particular, it says that $L$ *will* exhibit $\operatorname{Loc}(S)$ as a localization of $\mathcal C$ by the class of morphisms $L$ inverts—but these morphisms are in general a strict superclass of $S$. It also says that $L$ will have a universal property *similar* to the one for a localization by $S$, but which only applies to *left adjoint* functors that invert $S$.



Can the category of S-local objects be reflective but not a localization by S?
Not in general, no - there must be some additional conditions on $S$, such as a saturation condition.


Consider for instance the presentable case. Then if $S$ is small, $Loc(S) $ is always reflexive, and is always a localization of $C$ at the *saturated class generated by $S$*, but there is no reason to expect it to be the localization at $S$.


What's happening is that the class of arrows that are inverted by the left adjoint $C\to Loc(S)$ is always closed under colimits in $C$ (let $x$ be $S$-local, then the collection of arrows $f:y\to z$ such that $\hom(f,x)$ is an isomorphism is closed under colimits), so if $S$ is too small, they won't all be inverted by $C\to C[S^{-1}]$.


What you get in the presentable situation is exactly what you describe (because out of a presentable category, left adjoint = colimit preserving, you can replace "left adjoint" by "colimit preserving"): $C\to Loc(S)$ is the universal colimit preserving functor out of $C$ that inverts $S$.


It would be nice to have an explicit example here but of the top of my head I don't have one - a strategy to find one would be to find a functor $C\to D$ which is very very far from preserving colimits, but inverts $S$ nonetheless. Then there's a hope that it would invert $S$ but not the saturated class it generates, which would imply $C[S^{-1}]\not\simeq Loc(S)$


However, in general, model categories will provide good examples for another phenomenon : indeed for a model category $C$ with weak equivalences $W$, we have a very nice control over $C[W^{-1}]$ (this is why they were invented), but $Loc(W)$ will rarely be anything interesting, in fact it will often be trivial (in an appropriate sense), and so the inclusion does have a left adjoint, but $C[W^{-1}]$ is far from trivial.


Take for instance $C$ to be chain complexes over a ring $k$, and $W$ to be the class of quasi-isomorphisms (morphisms that induce an isomorphism in homology); then $C[W^{-1}]$ is very well known and well studied, it's $D(k)$, the derived category. In particular it is nonzero.


I claim that $Loc(W)$ consists only of the $0$ chain complex (in particular it is the trivial category $\*$, and so it is easy to check that the inclusion $Loc(W)\to C$ does have a left adjoint). Let $C\in Loc(W)$
Now consider the following chain complex : $D^n(k)= \dots \to 0\to k\to k\to 0\to \dots$ with only an identity arrow $k\to k$, the second $k$ in position $n-1$ and everything else $0$. Then the only map $D^n(k)\to 0$ is a quasi-isomorphism, so $0\to \hom(D^n(k),C)$ is an isomorphism. However $\hom(D^n(k),C)\cong C\_n$ because an arrow $D^n(k)\to C$ consists of $f:k\to C\_n$ and $g:k\to C\_{n-1}$ such that $\partial\_n f = g$, that is, an element $x\in C\_n$ and its image under $\partial\_n$, so in the end just an element $x\in C\_n$.


It follows that $C\_n=0$ so the claim now follows.


ADDED : Here's an explicit example of the first phenomenon : take the category of abelian groups, and the set $S$ to be the set of all $\mathbb Z/n\to 0$, where $n\geq 1$. Then $Loc(S)$ is the class of torsion-free abelian groups, and you have the classical unit $A\to A/tors(A)$ which shows that it is reflexive (although again it is a general fact, as $Ab$ is presentable). However, if you "naively" invert $S$, you don't get this. For instance, consider the $p$-Prüfer group $\mathbb Z/p^\infty = \mathrm{colim}\_n \mathbb Z/p^n$. This clearly dies in $Loc(S)$, but I claim it doesn't in $Ab[S^{-1}]$.


Indeed, consider the following category $C$ : objects are abelian groups, and $\hom(A,B)$ is the quotient of $\hom\_{Ab}(A,B)$ by the following subgroup : $f$ is in the subgroup if it factors through some finite abelian group. It's not hard to see that it does indeed define a category with the obvious composition law; and the functor $Ab\to C$ inverts $S$ (because in $\hom(\mathbb Z/n,\mathbb Z/n)$, $id - 0 \in$ the subgroup). So it factors as $Ab[S^{-1}]\to C$. But now $\mathbb Z/p^\infty$ is not isomorphic to $0$ in this new category, so it cannot be isomorphic to $0$ in $Ab[S^{-1}]$.


So what is happening here is exactly that we have this colimit of maps in $S$ ($\mathbb Z/p^n\to 0$) which is not in $S$ anymore. It must therefore be inverted by $C\to Loc(S)$, but it will not (in general) be inverted by $C\to C[S^{-1}]$, especially in cases where that colimit is "formal" (which it often is in presentable situations, almost by definition)


