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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 $i<b$? For example, let $K$ be centralized on degree $0$ and $b=0$, then right side must be zero in your notation..

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.

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

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.

Oops, yes I will fix this later. Thanks

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