The Stacks project

Lemma 13.36.2. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}'$ 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 $X' \to X \to Y \to X'[1]$ in $\mathcal{D}$ with $X'$ in $\mathcal{D}'$ and $\mathop{\mathrm{Hom}}\nolimits (A, Y) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$.

  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

\[ X'_1 \to X_1 \to Y_1 \to X'_1[1] \quad \text{and}\quad X'_2 \to X_2 \to Y_2 \to X'_2[1] \]

as in condition $P$. Since $\mathop{\mathrm{Hom}}\nolimits (X'_1, X'_2) = \mathop{\mathrm{Hom}}\nolimits (X'_1, X_2)$ by Lemma 13.36.1 there is a unique morphism $X'_1 \to X'_2$ such that the diagram

\[ \xymatrix{ X'_1 \ar[d] \ar[r] & X_1 \ar[d] \\ X'_2 \ar[r] & X_2 } \]

commutes. Choose an extension of this to a diagram

\[ \xymatrix{ X'_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & X'_1[1] \ar[d] \\ X'_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & X'_2[1] \ar[d] \\ X'_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & X'_3[1] \ar[d] \\ X'_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & X'_1[2] } \]

as in Proposition 13.4.22. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $\mathop{\mathrm{Hom}}\nolimits (A, Q_3) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ because this is true for $Q_1$ and $Q_2$ and because $\mathop{\mathrm{Hom}}\nolimits (A, -)$ is a homological functor (Lemma 13.4.2). Since $\mathcal{D}'$ is a full triangulated subcategory, we see that $X'_3$ is isomorphic to an object of $\mathcal{D}'$. 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 13.4.10. $\square$


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