Proposition 13.5.6. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Then there exists a unique structure of a pre-triangulated category on $S^{-1}\mathcal{D}$ such that $ \circ Q = Q \circ $ and the localization functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ is exact. Moreover, if $\mathcal{D}$ is a triangulated category, so is $S^{-1}\mathcal{D}$.

Proof. We have seen that $S^{-1}\mathcal{D}$ is an additive category and that the localization functor $Q$ is additive in Homology, Lemma 12.8.2. It follows from MS5 that there is a unique additive auto-equivalence $ : S^{-1}\mathcal{D} \to S^{-1}\mathcal{D}$ such that $Q \circ  =  \circ Q$ (equality of functors); we omit the details. We say a triangle of $S^{-1}\mathcal{D}$ is distinguished if it is isomorphic to the image of a distinguished triangle under the localization functor $Q$.

Proof of TR1. The only thing to prove here is that if $a : Q(X) \to Q(Y)$ is a morphism of $S^{-1}\mathcal{D}$, then $a$ fits into a distinguished triangle. Write $a = Q(s)^{-1} \circ Q(f)$ for some $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Choose a distinguished triangle $(X, Y', Z, f, g, h)$ in $\mathcal{D}$. Then we see that $(Q(X), Q(Y), Q(Z), a, Q(g) \circ Q(s), Q(h))$ is a distinguished triangle of $S^{-1}\mathcal{D}$.

Proof of TR2. This is immediate from the definitions.

Proof of TR3. Note that the existence of the dotted arrow which is required to exist may be proven after replacing the two triangles by isomorphic triangles. Hence we may assume given distinguished triangles $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ of $\mathcal{D}$ and a commutative diagram

$\xymatrix{ Q(X) \ar[r]_{Q(f)} \ar[d]_ a & Q(Y) \ar[d]^ b \\ Q(X') \ar[r]^{Q(f')} & Q(Y') }$

in $S^{-1}\mathcal{D}$. Now we apply Categories, Lemma 4.27.10 to find a morphism $f'' : X'' \to Y''$ in $\mathcal{D}$ and a commutative diagram

$\xymatrix{ X \ar[d]_ f \ar[r]_ k & X'' \ar[d]^{f''} & X' \ar[d]^{f'} \ar[l]^ s \\ Y \ar[r]^ l & Y'' & Y' \ar[l]_ t }$

in $\mathcal{D}$ with $s, t \in S$ and $a = s^{-1}k$, $b = t^{-1}l$. At this point we can use TR3 for $\mathcal{D}$ and MS6 to find a commutative diagram

$\xymatrix{ X \ar[r] \ar[d]^ k & Y \ar[r] \ar[d]^ l & Z \ar[r] \ar[d]^ m & X \ar[d]^{g} \\ X'' \ar[r] & Y'' \ar[r] & Z'' \ar[r] & X'' \\ X' \ar[r] \ar[u]_ s & Y' \ar[r] \ar[u]_ t & Z' \ar[r] \ar[u]_ r & X' \ar[u]_{s} }$

with $r \in S$. It follows that setting $c = Q(r)^{-1}Q(m)$ we obtain the desired morphism of triangles

$\xymatrix{ (Q(X), Q(Y), Q(Z), Q(f), Q(g), Q(h)) \ar[d]^{(a, b, c)} \\ (Q(X'), Q(Y'), Q(Z'), Q(f'), Q(g'), Q(h')) }$

This proves the first statement of the lemma. If $\mathcal{D}$ is also a triangulated category, then we still have to prove TR4 in order to show that $S^{-1}\mathcal{D}$ is triangulated as well. To do this we reduce by Lemma 13.4.15 to the following statement: Given composable morphisms $a : Q(X) \to Q(Y)$ and $b : Q(Y) \to Q(Z)$ we have to produce an octahedron after possibly replacing $Q(X), Q(Y), Q(Z)$ by isomorphic objects. To do this we may first replace $Y$ by an object such that $a = Q(f)$ for some morphism $f : X \to Y$ in $\mathcal{D}$. (More precisely, write $a = s^{-1}f$ with $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Then replace $Y$ by $Y'$.) After this we similarly replace $Z$ by an object such that $b = Q(g)$ for some morphism $g : Y \to Z$. Now we can find distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$ in $\mathcal{D}$ (by TR1), and morphisms $a : Q_1 \to Q_2$ and $b : Q_2 \to Q_3$ as in TR4. Then it is immediately verified that applying the functor $Q$ to all these data gives a corresponding structure in $S^{-1}\mathcal{D}$ $\square$

Comment #3572 by Herman Rohrbach on

Typos: 1. In the first sentence, the comma at the end of "is additive in Homology, Lemma 05QE," should be a period. 2. In the proof of TR1, "$a$ fits into a distinguish triangle" should be "$a$ fits into a distinguished triangle".

Comment #8343 by on

I think the translations $[n]:S^{-1}\mathcal{D}\to S^{-1}\mathcal{D}$ are not uniquely determined, and uniqueness is only guaranteed if we ask that $Q(X)=Q(X)$ holds (as actual equality, not natural isomorphism). Maybe the result should be restated to “there is a pre-triangulated structure on $S^{-1}\mathcal{D}$ that is unique up to (unique) exact isomorphism that turns $Q$ into a triangulated functor.”

I'm not sure if the following is completely right, but here the non-uniqueness argument I came up with: Suppose $F:S^{-1}\mathcal{D}\to S^{-1}\mathcal{D}$ is an additive isomorphism for which there are natural isomorphisms $\eta: F\cong \text{id}_{S^{-1}\mathcal{D}}$ and $\varepsilon: F^{-1}\cong \text{id}_{S^{-1}\mathcal{D}}$. Then, if we define $[n]'=F\circ [n]\circ F^{-1}$ as a new translation in $S^{-1}\mathcal{D}$, we get an isomorphism of functors $\eta_{[n]}\star\varepsilon :[n]'\cong [n]$ (notation from 4.28). For a triangle $\Delta=(X,Y,Z,f,g,h)$ in $S^{-1}\mathcal{D}$, define $\Delta'=(X,Y,Z,f,g,(\eta_{}\star\varepsilon)_X\circ h)$. If $\mathcal{T}$ is the class of distinguished triangles in $S^{-1}\mathcal{D}$, define $\mathcal{T}'=\{\Delta'\mid\Delta\in\mathcal{T}\}$. Then $Q:\mathcal{D}\to S^{-1}\mathcal{D}$ is also an exact functor of pre-triangulated categories when the codomain is the pre-triangulated category $(S^{-1}\mathcal{D},[\ ]',\mathcal{T}')$.

Comment #8346 by on

Update: I tried to get the uniqueness right in the pull request I just did on the GitHub project.

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