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Let $S$ be a class of maps of a category $\mathcal{C}$. The localization of $\mathcal{C}$ at $S$ is reflective when the localization functor $\mathcal{C} \to \mathcal{C}[S^{-1}]$ has a fully faithful right adjoint.

Question 1:

Are there any general conditions on $S$ which imply that $\mathcal{C}[S^{-1}]$ is reflective?

Suppose that $\mathcal{C}$ is locally presentable and that $S$ is a small set. Then the full subcategory $\mathcal{C}$ consisting of $S$-local objects is reflective. Let $\widetilde{S}$ be the class of $S$-local maps. Then $S \subseteq \widetilde{S}$ and $\mathcal{C}[S^{-1}]$ is reflective if and only if $\mathcal{C}[S^{-1}] = \mathcal{C}[\widetilde{S}^{-1}]$.

Question 2:

What are natural/useful examples of non-reflective localizations $\mathcal{C}[S^{-1}]$ (where $\mathcal{C}$ is locally presentable and $S$ is a set)?

The localization functor $\mathcal{C} \to \mathcal{C}[\widetilde{S}^{-1}]$ satisfies a modified version of the universal property of $\mathcal{C}[S^{-1}]$. Namely, we require all functors in the definition of a localization to be left adjoints. This definition makes sense for an arbitrary category $\mathcal{C}$ and an arbitrary class of maps $S$.

Question 3:

Is there a name for functors satisfying this universal property? Does it appear in the literature?

I want to call such a functor a reflective localization of $\mathcal{C}$ at $S$, but this terminology clashes with the ordinary notion of a reflective localization. Let us call it a quasireflective localization at $S$. Every reflective localization at $S$ is a quasireflective localization at $S$ by this proposition, but a quasireflective localization at $S$ (if it exists) is a reflective localization at $\widetilde{S}$ rather than at $S$. Thus, we can ask a weaker version of Question 2:

Question 2':

What are examples of quasireflective localizations at $S$ which are not reflective localizations at $S$.

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    $\begingroup$ As a natural example of non-reflective localization take $C=\mathrm{Ch}(\mathcal A)$, the chain complexes over a Grothendieck category, and $\mathcal S$, the class of quasi-isomorphisms. The corresponding localization is the derived category of $\mathcal A$ which is almost never contained in $\mathrm{Ch}(\mathcal A)$ as a full subcategory. $\endgroup$
    – Simone Virili
    Commented Sep 20, 2018 at 15:55
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    $\begingroup$ If $\mathcal A$ is a Grothendieck category, then $\mathrm{Ch}(\mathcal A)$ is locally $\alpha$-presentable (for a suitable choice of $\alpha$) and you can take $\mathcal S$ to be just the (small) set of quasi-isomorphisms among complexes where each term is $\alpha$-small. $\endgroup$
    – Simone Virili
    Commented Sep 20, 2018 at 15:58
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    $\begingroup$ @SimoneVirili Ah, indeed. I tihnk it is true more generally that the localization of many natural combinatorial model categories is not reflective. $\endgroup$
    – Valery Isaev
    Commented Sep 20, 2018 at 16:13

2 Answers 2

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As a consequence of the main theorem in

Di Liberti, I, and Loregian F.. "Homotopical algebra is not concrete." Journal of Homotopy and Related Structures (2017): 1-15.

(arXiv free version here), that the homotopy category of "almost all" known model categories is not concrete, I claim that "almost all" model categories produce non-reflective localizations.

Proof. All the examples we give in the paper are concrete categories (as a consequence of Remark 2.2, being concrete is a very weak assumption on a category); if you now assume $Ho(M)$ is a reflective localization of $M$, composing with a faithful functor $M \to Set$, you would get a faithful functor $Ho(M) \to M\to Set$.

This contradicts our Theorem 4.8.

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My answer comments questions Q1 and Q2 and is connected to the fact that Fosco gave an answer to the case in which $\mathcal{K}[S^{-1}]$ is not concrete. I will try to cover the other case, when $\mathcal{K}[S^{-1}]$ is concrete. I claim that this is a very strong requirement on $S$.

When a cocomplete category is concrete, quite often it is also locally presentable (Top is the most famous counterexample to this statement, which is still quite true). Thus, imagine to have a locally presentable $\mathcal{K}[S^{-1}]$ for which such a faithful right adjoint exist.

$$L: \mathcal{K} \leftrightarrows \mathcal{K}[S^{-1}]: i $$

Since both the categories are locally presentable, $i$ must be accessible for a big enough cardinal (Thm 2.23 LPAC).

Thm. An accessibly embedded reflective subcategory of a locally presentable category $\mathcal{A} \hookrightarrow \mathcal{K}$ is always of the form $\mathcal{K}[S^{-1}]$ for a small set $S$.

This result appears as Thm 1.39 in Locally Presentable and Accessible categories.

This means that quite often, the requirement for the existence of the faithful right adjoint is a strong requirement on the category $\mathcal{K}[S^{-1}]$, which in fact has to be equivalent to $\mathcal{K}[\bar{S}^{-1}]$ where $\bar{S}$ is (essentially) small.

May I conclude even that $\bar{S} \subset S$? Probably yes, in concrete examples $\bar{S}$ can even be chosen to lie inside $S$, this means that most of the information contained in $S$ is superfluous. This is due to the structure of the proof of 1.39.

All in all, putting together this answer with the other, the motto (which has no ambition of being a theorem) is: either $S$ is small or $\mathcal{K}[S^{-1}]$ is too big.

Of course, this super-general motto, has many counterexample. The easiest one is when $\mathcal{K}$ is connected and $S$ is everything, in this case $\mathcal{K}[S^{-1}] \cong 1$, which is a quite small category. Thus, one should be careful when applying this motto, that still in real life examples will turn out to be quite true.

This argument can even be strengthened under Vopenka principle. In fact, under Vopenka every reflective subcategory of a locally presentable category must be locally presentable (Cor 6.24 LPAC), thus the illegitimate upgrade from concrete to locally presentable comes for free.

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