Lemma 13.40.2. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ 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. $Z$ is in $\mathcal{A}^\perp$, and

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

Proof. By Lemma 13.4.2 the functor $\mathop{\mathrm{Hom}}\nolimits (A, -)$ is homological and hence we get a long exact sequence as in (13.3.5.1). Assume (1) and let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Then we consider the exact sequence

$\mathop{\mathrm{Hom}}\nolimits (A[1], Z) \to \mathop{\mathrm{Hom}}\nolimits (A, X) \to \mathop{\mathrm{Hom}}\nolimits (A, Y) \to \mathop{\mathrm{Hom}}\nolimits (A, Z)$

Since $A[1] \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we see that the first and last groups are zero. Thus we get (2). Assume (2) and let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Then we consider the exact sequence

$\mathop{\mathrm{Hom}}\nolimits (A, X) \to \mathop{\mathrm{Hom}}\nolimits (A, Y) \to \mathop{\mathrm{Hom}}\nolimits (A, Z) \to \mathop{\mathrm{Hom}}\nolimits (A[-1], X) \to \mathop{\mathrm{Hom}}\nolimits (A[-1], Y)$

and we conclude that $\mathop{\mathrm{Hom}}\nolimits (A, Z) = 0$ as desired. $\square$

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