The Stacks project

Lemma 12.8.3. Let $\mathcal{C}$ be an additive category. Let $S$ be a multiplicative system. Let $X$ be an object of $\mathcal{C}$. The following are equivalent

  1. $Q(X) = 0$ in $S^{-1}\mathcal{C}$,

  2. there exists $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that $0 : X \to Y$ is an element of $S$, and

  3. there exists $Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ such that $0 : Z \to X$ is an element of $S$.

Proof. If (2) holds we see that $0 = Q(0) : Q(X) \to Q(Y)$ is an isomorphism. In the additive category $S^{-1}\mathcal{C}$ this implies that $Q(X) = 0$. Hence (2) $\Rightarrow $ (1). Similarly, (3) $\Rightarrow $ (1). Suppose that $Q(X) = 0$. This implies that the morphism $f : 0 \to X$ is transformed into an isomorphism in $S^{-1}\mathcal{C}$. Hence by Categories, Lemma 4.27.21 there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves (1) $\Rightarrow $ (3). Similarly, (1) $\Rightarrow $ (2). $\square$


Comments (2)

Comment #332 by arp on

Typo: In the statement of the Lemma, it shouldn't say "compatible with the triangulated structure."

Comment #468 by Nuno on

Also, should be: Let be an object of .

There are also:

  • 3 comment(s) on Section 12.8: Localization

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Your email address will not be published.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 05QF. Beware of the difference between the letter 'O' and the digit '0'.