Lemma 13.39.4. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ be a full triangulated subcategory of $\mathcal{D}$. For an object $X$ of $\mathcal{D}$ consider the property $P(X)$: there exists a distinguished triangle $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $A$ in $\mathcal{A}$ and $B$ in $\mathcal{A}^\perp$.

1. If $X_1 \to X_2 \to X_3 \to X_1[1]$ is a distinguished triangle and $P$ holds for two out of three, then it holds for the third.

2. If $P$ holds for $X_1$ and $X_2$, then it holds for $X_1 \oplus X_2$.

Proof. Let $X_1 \to X_2 \to X_3 \to X_1[1]$ be a distinguished triangle and assume $P$ holds for $X_1$ and $X_2$. Choose distinguished triangles

$A_1 \to X_1 \to B_1 \to A_1[1] \quad \text{and}\quad A_2 \to X_2 \to B_2 \to A_2[1]$

as in condition $P$. Since $\mathop{\mathrm{Hom}}\nolimits (A_1, A_2) = \mathop{\mathrm{Hom}}\nolimits (A_1, X_2)$ by Lemma 13.39.2 there is a unique morphism $A_1 \to A_2$ such that the diagram

$\xymatrix{ A_1 \ar[d] \ar[r] & X_1 \ar[d] \\ A_2 \ar[r] & X_2 }$

commutes. Choose an extension of this to a diagram

$\xymatrix{ A_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & A_1[1] \ar[d] \\ A_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & A_2[1] \ar[d] \\ A_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & A_3[1] \ar[d] \\ A_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & A_1[2] }$

as in Proposition 13.4.23. By TR3 we see that $Q_1 \cong B_1$ and $Q_2 \cong B_2$ and hence $Q_1, Q_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $Q_3 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$ by Lemma 13.39.3. Since $\mathcal{A}$ is a full triangulated subcategory, we see that $A_3$ is isomorphic to an object of $\mathcal{A}$. Thus $X_3$ satisfies $P$. The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma 13.4.11. $\square$

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