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. $\square$

Comment #4000 by Zhiyu Zhang on

I am not sure this proposition is true... "Again using induction we see that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(H^ b(K)[-b], \tau _{\leq b - 1}K[1]) = \bigoplus \nolimits _{i < b} \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}^{b - i - 1}(H^ b(K), H^ i(K))$" shall the right side's index sum be $i\leq b$ rather than $i? For example, let $K$ be centralized on degree $0$ and $b=0$, 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 $\tau_{\leq b - 1}K[1]$ where it would have been more clear to write $(\tau_{\leq b - 1}K)[1]$, so first truncate and then shift to the left by $1$. Also, the truncation $\tau_{\leq b - 1}K$ has the same cohomology objects as $K$ in cohomological degree $\leq b - 1$ and zero in degrees $\geq b$. Some people would write $\tau^{\leq b - 1}K$ instead.

Comment #4003 by Zhiyu Zhang on

Oh I see, it's better to write $(\tau_{\leq b - 1}K)[1]$, thank you!

Comment #4005 by Zhiyu Zhang on

OK, It's also because of a typo that made me confused at first : the index $b-i-1$ on the right side of $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(H^ b(K)[-b], \tau _{\leq b - 1}K[1]) = \bigoplus \nolimits _{i < b} \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}^{b - i - 1}(H^ b(K), H^ i(K))$ shall be $b-i+1$, so $b-i+1 \geq 2$ 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.

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