Lemma 13.6.6. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. Set

13.6.6.1
$$\label{derived-equation-multiplicative-system} S = \left\{ \begin{matrix} f \in \text{Arrows}(\mathcal{D}) \text{ such that there exists a distinguished triangle } \\ (X, Y, Z, f, g, h) \text{ of }\mathcal{D}\text{ with } Z\text{ isomorphic to an object of }\mathcal{D}' \end{matrix} \right\}$$

Then $S$ is a multiplicative system compatible with the triangulated structure on $\mathcal{D}$. In this situation the following are equivalent

1. $S$ is a saturated multiplicative system,

2. $\mathcal{D}'$ is a saturated triangulated subcategory.

Proof. To prove the first assertion we have to prove that MS1, MS2, MS3 and MS5, MS6 hold.

Proof of MS1. It is clear that identities are in $S$ because $(X, X, 0, 1, 0, 0)$ is distinguished for every object $X$ of $\mathcal{D}$ and because $0$ is an object of $\mathcal{D}'$. Let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms contained in $S$. Choose distinguished triangles $(X, Y, Q_1, f, p_1, d_1)$, $(X, Z, Q_2, g \circ f, p_2, d_2)$, and $(Y, Z, Q_3, g, p_3, d_3)$. By assumption we know that $Q_1$ and $Q_3$ are isomorphic to objects of $\mathcal{D}'$. By TR4 we know there exists a distinguished triangle $(Q_1, Q_2, Q_3, a, b, c)$. Since $\mathcal{D}'$ is a triangulated subcategory we conclude that $Q_2$ is isomorphic to an object of $\mathcal{D}'$. Hence $g \circ f \in S$.

Proof of MS3. Let $a : X \to Y$ be a morphism and let $t : Z \to X$ be an element of $S$ such that $a \circ t = 0$. To prove LMS3 it suffices to find an $s \in S$ such that $s \circ a = 0$, compare with the proof of Lemma 13.5.3. 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 $(Q, Y, W, i, s, k)$. Here is a picture

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

Since $t \in S$ we see that $Q$ is isomorphic to an object of $\mathcal{D}'$. Hence $s \in S$. Finally, $s \circ a = s \circ i \circ g = 0$ as $s \circ i = 0$ by Lemma 13.4.1. We conclude that LMS3 holds. The proof of RMS3 is dual.

Proof of MS5. Follows as distinguished triangles and $\mathcal{D}'$ are stable under translations

Proof of MS6. Suppose given a commutative diagram

$\xymatrix{ X \ar[r] \ar[d]^ s & Y \ar[d]^{s'} \\ X' \ar[r] & Y' }$

with $s, s' \in S$. By Proposition 13.4.23 we can extend this to a nine square diagram. As $s, s'$ are elements of $S$ we see that $X'', Y''$ are isomorphic to objects of $\mathcal{D}'$. Since $\mathcal{D}'$ is a full triangulated subcategory we see that $Z''$ is also isomorphic to an object of $\mathcal{D}'$. Whence the morphism $Z \to Z'$ is an element of $S$. This proves MS6.

MS2 is a formal consequence of MS1, MS5, and MS6, see Lemma 13.5.2. This finishes the proof of the first assertion of the lemma.

Let's assume that $S$ is saturated. (In the following we will use rotation of distinguished triangles without further mention.) Let $X \oplus Y$ be an object isomorphic to an object of $\mathcal{D}'$. Consider the morphism $f : 0 \to X$. The composition $0 \to X \to X \oplus Y$ is an element of $S$ as $(0, X \oplus Y, X \oplus Y, 0, 1, 0)$ is a distinguished triangle. The composition $Y[-1] \to 0 \to X$ is an element of $S$ as $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle, see Lemma 13.4.11. Hence $0 \to X$ is an element of $S$ (as $S$ is saturated). Thus $X$ is isomorphic to an object of $\mathcal{D}'$ as desired.

Finally, assume $\mathcal{D}'$ is a saturated triangulated subcategory. Let

$W \xrightarrow {h} X \xrightarrow {g} Y \xrightarrow {f} Z$

be composable morphisms of $\mathcal{D}$ such that $fg, gh \in S$. We will build up a picture of objects as in the diagram below.

$\xymatrix{ & & Q_{12} \ar[rd] & & Q_{23} \ar[rd] \\ & Q_1 \ar[ld]_{\! + \! 1} \ar[ru] & & Q_2 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \ar[ru] & & Q_3 \ar[ld]_{\! + \! 1} \ar[ll]_{\! + \! 1} \\ W \ar[rr] & & X \ar[lu] \ar[rr] & & Y \ar[lu] \ar[rr] & & Z \ar[lu] }$

First choose distinguished triangles $(W, X, Q_1)$, $(X, Y, Q_2)$, $(Y, Z, Q_3)$ $(W, Y, Q_{12})$, and $(X, Z, Q_{23})$. Denote $s : Q_2 \to Q_1[1]$ the composition $Q_2 \to X[1] \to Q_1[1]$. Denote $t : Q_3 \to Q_2[1]$ the composition $Q_3 \to Y[1] \to Q_2[1]$. By TR4 applied to the composition $W \to X \to Y$ and the composition $X \to Y \to Z$ there exist distinguished triangles $(Q_1, Q_{12}, Q_2)$ and $(Q_2, Q_{23}, Q_3)$ which use the morphisms $s$ and $t$. The objects $Q_{12}$ and $Q_{23}$ are isomorphic to objects of $\mathcal{D}'$ as $W \to Y$ and $X \to Z$ are assumed in $S$. Hence also $s[1]t$ is an element of $S$ as $S$ is closed under compositions and shifts. Note that $s[1]t = 0$ as $Y[1] \to Q_2[1] \to X[2]$ is zero, see Lemma 13.4.1. Hence $Q_3[1] \oplus Q_1[2]$ is isomorphic to an object of $\mathcal{D}'$, see Lemma 13.4.11. By assumption on $\mathcal{D}'$ we conclude that $Q_3$ and $Q_1$ are isomorphic to objects of $\mathcal{D}'$. Looking at the distinguished triangle $(Q_1, Q_{12}, Q_2)$ we conclude that $Q_2$ is also isomorphic to an object of $\mathcal{D}'$. Looking at the distinguished triangle $(X, Y, Q_2)$ we finally conclude that $g \in S$. (It is also follows that $h, f \in S$, but we don't need this.) $\square$

Comment #328 by arp on

Typo: In the proof of MS3, second to last sentence, you write "Thus $j \circ f = j \circ g$ and we see that LMS3 holds," but $f$ and $g$ were never defined. You meant to write in the beginning that $f,g: X \rightarrow Y$ are morphisms and set $a = f-g$. Once you fix this you'll have the same problem with $g$ having two meanings as in my last comment, ha..

Comment #329 by arp on

Typo: In the sentence "Whence the morphism $Z' \to Z''$ is an element of $S$," it should say $Z \to Z'$. Also, really silly, but since I think you made a point of not assuming your categories are strictly full, maybe the previous sentence should say $Z''$ is isomorphic to an object of $\mathcal D'$.

Comment #1840 by Jonas on

1. Are the arrows leaving $Q_{12}$ and $Q_{23}$ in the nine object diagram going in the correct direction? The distinguished triangles $(Q_1, Q_{12}, Q_2)$ and $(Q_2, Q_{23}, Q_3)$ seem to give arrows in the opposite direction.

2. As $s[1]t = 0$ we conclude that $Q_3 \oplus Q_1[2]$ is an object of $\mathcal{D}'$. The distinguished triangle in question is $(Q_3, Q_1[2], P, 0, g, h)$, with $P$ isomorphic to $Q_3[1] \oplus Q_1[2]$, hence this should be the object in $\mathcal{D}'$.

Thanks for a nice proof.

Comment #1877 by on

Dear Jonas, yes in both cases. I also removed the top arrow between $Q_{12}$ and $Q_{23}$ because it serves no purpose and wasn't defined in the proof. See changes here.

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