Lemma 13.5.3. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Let

$S = \{ f \in \text{Arrows}(\mathcal{D}) \mid F(f)\text{ is an isomorphism}\}$

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 Lemma 13.4.3. 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 $(Z, X, Q, t, d, h)$ using TR1. Since $a \circ t = 0$ we see by Lemma 13.4.2 there exists a morphism $i : Q \to Y$ such that $i \circ d = a$. Finally, using TR1 again we can choose a triangle $(Q, Y, W, i, j, k)$. Here is a picture

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

OK, and now we apply the functor $F$ to this diagram. Since $t \in S$ we see that $F(Q) = 0$, see Lemma 13.4.9. Hence $F(j)$ is an isomorphism by the same lemma, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ d = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual. $\square$

Comment #327 by arp on

Typo: In the proof $g$ has two meanings, as a morphism $X \rightarrow Y$ and a morphism $X \rightarrow Q$ in a distinguished triangle, which is a little confusing. By the way in the sentence where the second meaning of $g$ is introduced, "Choose a distinguished triangle $(Z, X, Q, t, g, h)$ using TR1 and TR2," I think you're just using TR1 according to your definitions.

Comment #375 by Fan on

The last line: $j \circ f$ should = $j \circ g$ rather than $j \circ d$.

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