12
$\begingroup$

This is cross-posted from MSE (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 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 $S$.

$\endgroup$
14
$\begingroup$

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)

$\endgroup$
1
  • 1
    $\begingroup$ @sarahzrf : you might be interested by the (somewhat simpler) example I added of the first phenomenon I described :) $\endgroup$ – Maxime Ramzi Jan 7 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.