Lemma 13.40.3. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Consider a distinguished triangle

$X \to Y \to Z \to X[1]$

of $\mathcal{D}$. The following are equivalent

1. $X$ is in ${}^\perp \mathcal{B}$, and

2. $\mathop{\mathrm{Hom}}\nolimits (Y, B) = \mathop{\mathrm{Hom}}\nolimits (Z, B)$ for all $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$.

Proof. Dual to Lemma 13.40.2. $\square$

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