# Bousfield localization of triangulated categories:equivalent conditions

In these notes on pages 60-64 Daniel Murfet proves the equivalence of 6 conditions of what it means for the Verdier quotient to be Bousfield localization. I, however, do not understand certain steps in implications (1) $$\Longrightarrow$$ (2) and (6) $$\Longrightarrow$$ (1):

First, preliminary definitions:

Let $$T$$ be a triangulated category. A localization in $$T$$ is a pair $$(l,\eta)$$ where $$l\colon T\to T$$ is a triangulated functor and $$\eta\colon 1\Rightarrow l$$ is a trinatural transformation such that $$l(\eta_X) = \eta_{l(X)}$$ and this morphism $$l(X)\to l(l(X))$$ is an isomorphism.

It the notes it is proven that the kernel $$L$$ of $$l$$ is a thick localizing subcategory of $$T$$ and that the following are equivalent:

$$(1)$$ $$X \in L^{\bot}$$

$$(2)$$ $$\eta_X\colon X\to l(X)$$ is an isomorphism,

$$(3)$$ $$X \cong l(Y)$$ for some $$Y \in T$$.

As a corollary, we have $${\bot}_{(L^{\bot})} = L$$ where $$L^{\bot}$$ is the triangulated subcategory consiting of those objects $$X$$ for which $$Hom_T(Y,X)$$ is zero for all $$Y \in L$$ and $${\bot}_L$$ is a full subcategory consiting of those objects $$X$$ for which $$Hom_T(X,Y)$$ is zero for all $$Y \in L$$.

Now here goes the proposition:

Let $$T$$ be a triangulated category, $$L$$ a thick subcategory. Denote by $$i\colon L\to T$$ and $$j\colon L^{\bot}\to T$$ the inclusions and $$Q\colon T\to T/L$$ the Verdier quotient. Then the following are equivalent:

(1) There is a localization $$(l,\eta)$$ whose kernel is $$L$$,

(2) The functor $$Q$$ has a right adjoint,

(3) The composition $$Qj$$ is an equivalence of categories,

(4) The functor $$j$$ has a left adjoint and and $${\bot}_{(L^{\bot})} = L$$,

(5) The functor $$i$$ has a right adjoint,

(6) For every $$M \in T$$ there is a distinguished triangle $$N_M\to M\to B_M\to \Sigma N_M$$ with $$N_M \in L$$ and $$B_M \in L^{\bot}$$.

Here is a proof of (1) $$\Longrightarrow$$ (2)

What I don't understand here is why chosen $$g$$ is unique making the diagram commute.

The proof (6) $$\Longrightarrow$$ (1) is, unfortunately, quite long.

Here I don't understand several things.

First is, why $$\psi_M$$ and $$\phi_M$$ are isomorphisms (as Murfet claim "by symmetry"). It is known that if two of three morphisms consituting a morphism of triangles are isomorphisms, then so is the third one, but here we have only one known isomorphism (the identity).

Second, why do we need $$l^a$$ and $$u$$? They have nothing to do with the definition of a localization.

Third, why a morphism $$C \cong B_Z$$ compatible with $$j, v_Z$$ is an isomorphism?

Fourth, why the homotopy kernel of a $$L$$-localization must belong to $$L$$, and why from this it follows that the final diagram commutes?

Fifth and finally, why $$(l,v)$$ is localization. I understand why $$l(v_M) = v_{l(M)}$$ but I don't see why this morphism is an isomorphism.

I know that's a lot of questions, but I kinda got lost there in the end and cannot get "unlost" without further help. I'm sorry if this is frowned upon here.

I have to note that almost all questions for (6) $$\Longrightarrow$$ (1) except the fourth one would be easily resolved if $$v_M$$ would be an isomorphism, but it doesn't appear to be the case lest $$N_M = 0$$.

• For (1) $$\Rightarrow$$ (2), One uses the universal property of $$Q$$ to see that $$\ell$$ factors as $$R \circ Q$$, then it is basically formal to see that $$R$$ is a right adjoint by checking the triangular equations.
• As for (6) $$\Rightarrow$$ (1) one needs to check the functoriality of the triangle, a fact more or less formal in view of Lemma 1.4 in loc. cit. which in turns refers to Beilinson-Bernstein-Deligne-Gabbers's classic. The idea is that, in TR3 axiom of triangulated categories, if you have that $$\mathrm{Hom}(X,Z′[−1])=0$$ then the completion of the map of triangles is unique.