Are there adjoint localizations that aren't reflexive localizations?

89 Views Asked by At

If you have a category $C$ and a class of weak equivalences $W$ ( we may assume it has $2$-out-of-$3$, or even that it's saturated, it won't change the question), you can localize (modulo set theory) at $W$ : $\gamma : C \to C[W^{-1}]$.

A lot of the time, this is actually a reflexive localization : that is, $\gamma$ has a fully faithful right adjoint which exhibits $C[W^{-1}]$ as the full subcategory of $C$ on $W$-local objects, and $\gamma$ is then just a reflector on that full subcategory.

My question is about whether this situation is necessary given that $\gamma$ has a right adjoint.

More precisely : assume $\gamma$ (which is a localization functor) has a right adjoint $i$. Is it necessarily fully faithful ?

Some thoughts on this : if $C$ has enough $W$-local objects (that is, every $x$ has a weak equivalence to a $W$-local object), then we are in the situation of a reflexive localization.

Another thing is that if the unit of the adjunction is a natural weak equivalence, it also follows from some simple computations that we are dealing with a reflexive localization.

Apart from that I haven't really made progress on the question. I know that for instance there can be a full subcategory $D$ such that $\gamma_{\mid D}$ is an equivalence without $\gamma$ having a right adjoint; I also know if there is such a full subcategory, the inclusion $D\to C$ can have a left adjoint without it being induced by $\gamma_{\mid D}^{-1} \circ \gamma$. Finally, if there is such a full subcategory $D$, every object in $D$ is $W$-local if and only if the localization is reflexive.

1

There are 1 best solutions below

5
On BEST ANSWER

This is true, although it seems to be difficult to give a straightforward argument. Here is a proof that I learned from Cisinski's book in the $\infty$-categorical case.

Suppose that $\gamma$ admits a right adjoint $i$. The precomposition functor $\gamma^*:\mathrm{Set}^{C[W^{-1}]^{\mathrm{op}}}\to \mathrm{Set}^{C^{\mathrm{op}}}$ is fully faithful, by the universal property of localization-this would be true with any category in place of $\mathrm{Set}$. (We'd better assume $\mathrm{Set}$ is big enough that $C[W^{-1}]$ looks small to it.)

Since mapping into $\mathrm{Set}$ is a 2-functor $\mathrm{Cat}^{\mathrm{op}}\to \mathrm{Cat}$, while $(-)^\mathrm{op}$ is a 2-functor $\mathrm{Cat}^{\mathrm{co}}\to \mathrm{Cat}$, the 2-functor $\gamma\mapsto \gamma^*$ is contravariant at both levels: it goes from $\mathrm{Cat}^{\mathrm{op-co}}\to \mathrm{Cat}$. The goal here is to note that we still have $\gamma\dashv i$ in $\mathrm{Cat}^{\mathrm{op-co}}$.

This means that $\gamma^*\dashv i^*$, so that $\gamma^*$ may be naturally identified with $i_!$, that is, with left Kan extension along $i$. Finally, since $i_!$ is fully faithful, then $i$ must be as well. This is simply because $i_!\left(C[W^{-1}](-,c)\right)=C(-,i(c))$.

Note that, since the opposite of a localization is a localization, the same result holds if $\gamma$ has a left adjoint.