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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)

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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.

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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 $N_M = 0$.

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The theorem in Murfet's notes comes form the paper "Localization in Categories of Complexes and Unbounded Resolutions", Canad. J. Math. Vol. 52 (2), 2000

It corresponds to Proposition 1.6.

  • 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.

Note that a localizing subcategory is always thick by Eilenberg's swindle. See the Remark at the end of page 227 in the cited paper.

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  • $\begingroup$ Thanks you for attention to this question, but I'm fine with the idea and I know the proof of 1.4, it's the respective details I'm having problems with. $\endgroup$
    – Jxt921
    Jan 11 at 20:03
  • $\begingroup$ One of my problems is that I understand why the triangle is functorial, but I don't see why the functor it is exact. $\endgroup$
    – Jxt921
    Jan 11 at 20:11
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    $\begingroup$ At first glance, it should follow form BBD, Propostion 1.1.11, I guess. $\endgroup$
    – Leo Alonso
    Jan 11 at 20:13

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