This question is a follow-up to [this question](https://mathoverflow.net/questions/215802/vanishing-natural-transformation-and-strong-generator) I asked some time ago. Let $X$ be a **smooth** projective variety of dimension $n$ over $\mathbb{C}$. Let $\omega \in H^{n}(X,K\_X)$, $\omega \neq 0$. Let 
$$A \longrightarrow B \longrightarrow C \longrightarrow A[1]$$


be an exact triangle in $D^b(X)$. Assume that the morphism:
$$\mathrm{id}\_A \otimes \omega : A \longrightarrow A \otimes K\_X[n]$$
vanishes in $\mathrm{Hom}(A,A\otimes K\_X[n])$ and that the morphism:
$$\mathrm{id}\_B \otimes \omega : B \longrightarrow B \otimes K\_X [n]$$
vanishes in $\mathrm{Hom}(B,B\otimes K\_X[n])$. I'd like to know if the morphism:
$$\mathrm{id}\_C \otimes \omega : C \longrightarrow C \otimes K\_X[n]$$
vanishes in $\mathrm{Hom}(C,C\otimes K\_X[n])$?


I know that there is no *five lemma* for zero-morphisms and that if this statement is true it is not just a formal consequence of the definition of a triangulated category (see the answer to the question I am referring to in the intro). 


On the other hand, seen the very special nature of the involved morphisms, I tought it might be true. Or perhaps this is plain wrong and a counter-example would be greatly appreciated.


Thanks in advance!



Vanishing natural transformation exact triangle
So I think I have an argument which shows that if $\mathrm{id}\_{A} \otimes \omega : A \longrightarrow A \otimes K\_X[n]$ is zero and $\mathrm{id}\_{B} \otimes \omega : B \longrightarrow B \otimes K\_X[n]$ is zero then $\mathrm{id}\_{C} \otimes \omega : C \longrightarrow C \otimes K\_X[n]$ is not necessarily zero. This argument uses crucially the result of Thomason that was brought to my attention by Jeremy Rickard (who I heartily thank for that).


Let $X$ be a K3 surface and let $\mathcal{T}$ be the cone of the natural map:
$$\mathcal{O}\_X \longrightarrow \mathcal{O}\_X[2]$$
induced by $\omega$. The object $\mathcal{T}$ has rank $0$ in $D^b(X)$ and one easily computes that $\mathrm{End}(\mathcal{T}, \mathcal{T}) = \mathbb{C}$. Assume that the morphism $\mathrm{id}\_{\mathcal{T}} \otimes \omega : \mathcal{T} \longrightarrow \mathcal{T}[2]$ is not zero. Then it is a generator of $\mathrm{Ext}^2(\mathcal{T},\mathcal{T}) = \mathrm{End}(\mathcal{T})^\* = \mathbb{C}$. This is impossible. Indeed, we have $$\mathrm{Trace}(\mathrm{id}\_{\mathcal{T}} \otimes \omega) = \mathrm{rank}(\mathcal{T}).\omega = 0$$ but Serre duality implies that:
$$\mathrm{Trace} : \mathrm{Ext}^2(\mathcal{T}, \mathcal{T}) \longrightarrow H^2(X,\mathcal{O}\_X)$$
is surjective. This shows that $\mathrm{id}\_{\mathcal{T}} \otimes \omega = 0$ in $\mathrm{Ext}^{2}(\mathcal{T}, \mathcal{T})$.


Now, assume that for any exact triangle $A \longrightarrow B \longrightarrow C$ in $D^b(X)$ such that $\mathrm{id}\_{A} \otimes \omega : A \longrightarrow A[2]$ is zero and $\mathrm{id}\_{B} \otimes \omega : B \longrightarrow B[2]$ is zero, we have $\mathrm{id}\_{C} \otimes \omega : C \longrightarrow C[2]$ is also zero. (I denote this hypothesis by $(\*)$).


Then, the subcategory of $D^b(X)$ consisting of objects $S$ such $\mathrm{id}\_{S} \otimes \omega$ is zero would be a thick $\otimes$-triangulated subcategory of $D^b(X)$ (see the comment of Jeremy Rickard to the question). 


It has been checked that $\mathcal{T}$ is in this category and the cohomological support of $\mathcal{T}$ is $X$. Hence, by Thomason classification of thick $\otimes$-triangulated subcategories of $D^b(X)$, we deduce that for any $S \in D^b(X)$ such that the cohomological support of $S$ is $X$, we have the vanishing of $\mathrm{id}\_{S} \otimes \omega : S \longrightarrow S \otimes[2]$. This is clearly absurd and hypothesis $(\*)$ is incorrect.


