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Suppose there are two classes of morphisms $w_1, w_2$ in $C$ and two two reflective localizations $L_1: C \overset{\rightarrow}{\hookleftarrow} C^\text{$w_1$-local}: i_1$ and $L_2: C \overset{\rightarrow}{\hookleftarrow} C^\text{$w_2$-local}: i_2$. I am wondering when $L_1$ restricts to an adjunction $$L_1: C^\text{$w_2$-local} \overset{\rightarrow}{\hookleftarrow} C^\text{$w_1$, $w_2$-local}.$$ I think this holds when if the endofunctors $i_1\circ L_1$ and $i_2 \circ L_2$ of $C$ commute.

Question: Is there an easily checkable criterion on $w_1$, $w_2$ to determine whether these endofunctors commute?

Another reasonable condition is to assume that for any morphism $f_1: X_1 \rightarrow Y_1$ in $w_1$, the objects $X_1$, $Y_1$ are $w_2$-local. In this case, $f_1: X_1 \rightarrow Y_1$ is a morphism in $C^\text{$w_2$-local}$, and hence we can form $(C^\text{$w_2$-local})^\text{$w_1$-local}$ (which I believe is equivalent to $C^\text{$w_1$, $w_2$-local}$) and produce the reflective localization $$ C^\text{$w_2$-local} \overset{\rightarrow}{\hookleftarrow} (C^\text{$w_2$-local})^\text{$w_1$-local} = C^\text{$w_1$, $w_2$-local}. $$ However, I am not sure that this localization is the restriction of $L_1$.

I also suspect that this problem might be a special case of a more general setting with pairs of adjoint functors that are not reflective localizations, i.e. a pair of adjoint functors whose composition endofunctors commute.

** I am thinking about this problem in the $\infty$-categorical setting but an answer in the setting of ordinary categories would also be helpful.

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  • $\begingroup$ Rather than generalise the problem I suggest you start by considering the simpler one of two idempotents (retracts) on a set. The intersection of the two fixed subsets need not be fixed unless there is some commutativity condition. However, a weaker condition such as $e\cdot f\cdot e=f\cdot e$ may suffice, but giving two isomorphic subsets. $\endgroup$ Commented Oct 30 at 16:57
  • $\begingroup$ Thanks. I mentioned commutativity in the title, but will make a more explicit reference in the post. $\endgroup$
    – user39598
    Commented Oct 30 at 17:17
  • $\begingroup$ A few thoughts -- (1.) The case of a recollement is somehow the "opposite extreme" where composing one way gives zero but composing the other way records enough info to recover the ambient category. Maybe one could try build up a general pair of localizations in terms of recollements and commuting pairs? (2.) If you impose conditions like "$L_i$ exact" or "$R_i$ has a further adjoint" things might simplify. (3.) Favorite example: Nisnevich / $\mathbb A^1$ localization don't commute... $\endgroup$ Commented Oct 30 at 17:20
  • $\begingroup$ ... localizations form a (4.) band (i.e. an idempotent semigroup) under composition. Commutative bands = semilattices. Varieties of bands are classified... (not sure what to do with that) (5.) example of (2) : if $R_1$ has a further adjoint (i.e. $L_1$ is smashing), then $L_1$ commutes with all other localizations (er -- in some contexts?). It's in the center of the band. (6.) Stuff on the Bousfield lattice or left-exact localizations (which form a distributive lattice (does that mean they commute?) -- see Borceux-Kelly) might be relevant. $\endgroup$ Commented Oct 30 at 17:23
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    $\begingroup$ Reflective localisations are idempotent monads. For monads, there is a notion of distributive law. As far as I can tell it does not degenerate in the idempotent case to commutativity of endofunctors. $\endgroup$
    – Zhen Lin
    Commented Oct 30 at 23:26

1 Answer 1

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Here is one general result in terms of $w_1$ and $w_2$ (or their saturations).

Suppose that $C$ is presentable, and that for any morphism $f: X_2\rightarrow Y_2$ in $w_2$, the objects $X_2, Y_2$ are $\overline{w_1}$-colocal, meaning that $Hom(X_2, f)$ and $Hom(Y_2, f)$ are isomorphisms for any morphism $f$ in the saturation $\overline{w_1}$ of $w_1$. Then $i_1 \circ L_1$ preserves $w_2$-local objects.

To see this, recall that we need to prove that $i_1\circ L_1$ preserves $w_2$-local objects: if $c$ is $w_2$-local, then $i_1\circ L_1(c)$ is also $w_2$-local. In the presentable case, the unit $\eta_c: c\rightarrow i_1 \circ L_1(c)$ is in the saturation $\overline{w_1}$ of $w_1$.

For any morphism $f: X_2\rightarrow Y_2$ in $w_2$ there is a commutative diagram $\require{AMScd}$ \begin{CD} Hom(Y_2, c) @>\eta_c>>Hom(Y_2, i_1 \circ L_1(c))\\ @V f V V @VV f V\\ Hom(X_2, c) @>\eta_c>> Hom(X_2, i_1 \circ L_1(c)) \end{CD} Since $c$ is $w_2$-local by assumption, the left vertical map is an equivalence. Since $\eta_c$ is in $\overline{w_1}$ and $X_2$ and $Y_2$ are $\overline{w_1}$-colocal, the top and bottom maps are also equivalences. Hence, the right vertical map is also an equivalence and hence $i_1 \circ L_1(c)$ is a $w_2$-local object, as desired.

Ideally, one wouldn't need to form $\overline{w_1}$ but only consider $w_1$. It is true that if an object $c$ is $w_1$-local, then $c$ is $\overline{w_1}$-local. However, I am not sure about colocal (due to the mix-and-match of localizations and colocalizations).

Question: If $c$ is $w_1$-colocal, then is $c$ $\overline{w_1}$-colocal?

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