Lemma 13.5.5. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor between a pre-triangulated category and an abelian category. Let

$S = \{ f \in \text{Arrows}(\mathcal{D}) \mid H^ i(f)\text{ is an isomorphism for all }i \in \mathbf{Z}\}$

Then $S$ is a saturated (see Categories, Definition 4.27.20) multiplicative system compatible with the triangulated structure on $\mathcal{D}$.

Proof. We have to prove axioms MS1 – MS6, see Categories, Definitions 4.27.1 and 4.27.20 and Definition 13.5.1. MS1, MS4, and MS5 are direct from the definitions. MS6 follows from TR3 and the long exact cohomology sequence (13.3.5.1). By Lemma 13.5.2 we conclude that MS2 holds. To finish the proof we have to show that MS3 holds. To do this let $f, g : X \to Y$ be morphisms of $\mathcal{D}$, and let $t : Z \to X$ be an element of $S$ such that $f \circ t = g \circ t$. As $\mathcal{D}$ is additive this simply means that $a \circ t = 0$ with $a = f - g$. Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2. Since $a \circ t = 0$ we see by Lemma 13.4.2 there exists a morphism $i : Q \to Y$ such that $i \circ g = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture

$\xymatrix{ Z \ar[r]_ t & X \ar[r]_ g \ar[d]^1 & Q \ar[r] \ar[d]^ i & Z[1] \\ & X \ar[r]_ a & Y \ar[d]^ j \\ & & W }$

OK, and now we apply the functors $H^ i$ to this diagram. Since $t \in S$ we see that $H^ i(Q) = 0$ by the long exact cohomology sequence (13.3.5.1). Hence $H^ i(j)$ is an isomorphism for all $i$ by the same argument, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ g = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual. $\square$

Comments (2)

Comment #376 by Fan on

Well, this is mostly repetition of the previous lemma. Can it be deduced by setting $\mathcal D' = Ch(A)$, and $F = H$ in the previous lemma?

Comment #385 by on

@#376: First of all, I don't think it can be deduced in that way. Try it!

General remark: Often it does make sense to repeat very similar arguments, because then you see what is the key thing you have to change. We can do this also because we are not writing a paper, and hence there is no restriction on the number of pages. But of course, if you get many similar arguments, then it makes sense to set up a kind of "machine" that does all of them at once.

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• 3 comment(s) on Section 13.5: Localization of triangulated categories

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