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

$Z$ is in $\mathcal{A}^\perp $, and

$\mathop{\mathrm{Hom}}\nolimits (A, X) = \mathop{\mathrm{Hom}}\nolimits (A, Y)$ 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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