The Stacks project

Lemma 13.27.8. Let $\mathcal{A}$ be an abelian category. Assume $\mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {A}(B, A) = 0$ for any pair of objects $A$, $B$ of $\mathcal{A}$. Then any object $K$ of $D^ b(\mathcal{A})$ is isomorphic to the direct sum of its cohomologies: $K \cong \bigoplus H^ i(K)[-i]$.

Proof. Choose $a, b$ such that $H^ i(K) = 0$ for $i \not\in [a, b]$. We will prove the lemma by induction on $b - a$. If $b - a \leq 0$, then the result is clear. If $b - a > 0$, then we look at the distinguished triangle of truncations

\[ \tau _{\leq b - 1}K \to K \to H^ b(K)[-b] \to (\tau _{\leq b - 1}K)[1] \]

see Remark 13.12.4. By Lemma 13.4.10 if the last arrow is zero, then $K \cong \tau _{\leq b - 1}K \oplus H^ b(K)[-b]$ and we win by induction. Again using induction we see that

\[ \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(H^ b(K)[-b], (\tau _{\leq b - 1}K)[1]) = \bigoplus \nolimits _{i < b} \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}^{b - i + 1}(H^ b(K), H^ i(K)) \]

Since $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A) = 0$ for $i \geq 2$ and any pair of objects $A, B$ of $\mathcal{A}$ by our assumption and Lemma 13.27.7 we are done. $\square$


Comments (6)

Comment #4000 by Zhiyu Zhang on

I am not sure this proposition is true... "Again using induction we see that " shall the right side's index sum be rather than ? For example, let be centralized on degree and , then right side must be zero in your notation..

Comment #4001 by on

This lemma is definitively true and the proof is correct too. The problem may be that we use the notation where it would have been more clear to write , so first truncate and then shift to the left by . Also, the truncation has the same cohomology objects as in cohomological degree and zero in degrees . Some people would write instead.

Comment #4003 by Zhiyu Zhang on

Oh I see, it's better to write , thank you!

Comment #4005 by Zhiyu Zhang on

OK, It's also because of a typo that made me confused at first : the index on the right side of shall be , so and we can use the assumption to deduce this lemma.

Comment #4006 by on

Oops, yes I will fix this later. Thanks

Comment #4117 by on

Dear Zhiyu Zhang, thanks again for pointing out this very confusing mistake. I have now finally fixed this here.


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Your email address will not be published.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