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. $\square$

## Comments (0)

There are also:

• 3 comment(s) on Section 13.40: Admissible subcategories

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0H0M. Beware of the difference between the letter 'O' and the digit '0'.