# a zero morphism on t-structures

Good morning to everyone, I am writing here because I need to understand better some topics about t-structures on triangulated categories.

Consider this statement: take a, b in $$\mathbb{Z}$$, $$(\mathscr{C}^{\leq 0}, \mathscr{C}^{\geq 0})$$ a t-structure on a triangulated category $$(\mathscr{C}, [1], \partial)$$ and $$C$$ an object of $$\mathscr{C}$$.

For any $$n$$ in $$\mathbb{Z}$$, $$X$$ in $$Ob(\mathscr{C})$$, I call

$$\Delta_{X}^{0} : \tau^{\leq 0}(X) \rightarrow^{T^{\leq 0}(X)} X \rightarrow^{T^{\geq 1}(X)} \tau^{\geq 1}(X) \rightarrow^{h_{X}^{0}} \tau^{\leq 0}(X)[1]$$

the distinguished triangle coming from the axioms of the t-structure.

Is the equality

$$T^{\geq a}(C) \circ T^{\leq b}(C) = 0$$

true if $$b \lneq a$$?

I tried to prove it, but I did not manage. Thank you in advance.

• Couldn't you reduce to the case $a = 1$, $b = 0$ in which case this is one of the axioms for a t-structure?
– JHF
Jan 14 at 20:43
• Did you check it using induction on a - b? By the definition of the Tn's built by the T0 and T1. (Sorry for my bad English) Anyway I thank @JHF , for the quick hint. Jan 15 at 13:06
• My thought was to use the shift equivalences to shift $b$ to $0$, and then observe that the truncation factors through $T^{\geq 1}$. But I suppose the idea is not that different from yours.
– JHF
2 days ago

By the axioms

$$\bullet$$ $$\mathscr{C}^{\leq -1} \subseteq \mathscr{C}^{\leq 0}$$

$$\bullet$$ $$\mathscr{C}^{\geq 1} \subseteq \mathscr{C}^{\geq 0}$$

$$\bullet$$ $$Hom_{\mathscr{C}}(\mathscr{C}^{\leq 0}, \mathscr{C}^{\geq 1}) \cong 0$$

I can prove that

$$\hspace{-130pt} Hom_{\mathscr{C}}(\mathscr{C}^{\leq n}, \mathscr{C}^{\geq m}) \cong 0$$

for any $$n$$ and $$m$$ integers such that $$n \lneq m$$, using induction on $$m - n \in \mathbb{N}_{\geq 1}$$.

By the relation $$\hspace{-90pt} T^{\geq a}(C) \circ T^{\leq b}(C) \in Hom_{\mathscr{C}}(\mathscr{C}^{\leq b}, \mathscr{C}^{\geq a}) \cong 0$$ I am done.

If there are some mistakes, please let me know.