# ct.category theory – 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 $$Cto Loc(S)$$ is always closed under colimits in $$C$$ (let $$x$$ be $$S$$-local, then the collection of arrows $$f:yto z$$ 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 $$Cto C(S^{-1})$$.

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”): $$Cto Loc(S)$$ 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 $$Cto D$$ 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 $$C(S^{-1})notsimeq Loc(S)$$

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 $$Cin Loc(W)$$
Now consider the following chain complex : $$D^n(k)= dots to 0to kto kto 0to dots$$ with only an identity arrow $$kto k$$, 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 $$0to hom(D^n(k),C)$$ is an isomorphism. However $$hom(D^n(k),C)cong C_n$$ because an arrow $$D^n(k)to C$$ consists of $$f:kto C_n$$ and $$g:kto C_{n-1}$$ such that $$partial_n f = g$$, that is, an element $$xin C_n$$ and its image under $$partial_n$$, so in the end just an element $$xin C_n$$.

It follows that $$C_n=0$$ so the claim now follows.