**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 13.40.2 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 13.40.4. 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.
$\square$

## Comments (0)

There are also: