Does $\operatorname{Hom}(A, B)=0$ imply $\operatorname{Hom}(A, B[1])=0$ in a triangulated category?

Question

Let $D$ be a triangulated category and $A, B \in D$. Then, in generally, does the condition $\operatorname{Hom}(A, B)=0$ yields $\operatorname{Hom}(A, B[i])=0$?

If the claim is right, how to proof?

Answer 1

No. For example, if we are working in $D^b(Coh(\mathbb{P}^1))$ and take $A = \mathcal{O}_p$ the skyscraper sheaf of a point and $B = \mathcal{O}$, then $Hom_{D^b}(A,B)= Hom_{Sh}(A,B) = 0$ since $\mathcal{O}$ is torsion-free, while $Hom_{D^b}(A,B[1]) = Ext^1(A,B) \neq 0$ since $Ext^1(\mathcal{O}_p, \mathcal{O}) \cong Hom(\mathcal{O}, \mathcal{O}_p)^* \neq 0$ by Serre duality.

Answer 2

Certainly not. The simplest triangulated categories to think about are the categories of chain complexes of objects of some abelian category and homotopy classes of maps. For instance, think about maps in this categorified between chain complexes with only one nonzero object, in the $i$th dimension, denoted $A[i]$. Then there are never any nonzero maps $A[i]\to B[j]$ unless $i=j$, while the maps $A[i]\to B[i]$ are just the maps $A\to B$.

This is the simplest example but not the most characteristic one, which would be the derived category of an abelian category. In there, the maps $A[i]\to B[j]$ correspond to $\mathrm{Ext}^{j-i}(A,B)$, at least as long as the ext functors exist. And there are certainly choices if $A$ and $B$ with arbitrary sets of exts vanishing and nonvanishing. Most simply, remember that there are no homs from $\mathbb Z/2$ to $\mathbb Z$, but there are exts!

Answer 3

Fix a field $k$, then category of $\mathbb Z$-graded $k$-vector spaces (with degree zero homogeneous maps) is triangulated. The shifting is the usual one, and the mapping cone of a map $f:X\to Y$ is $Ker(f)[1]\oplus Y/Im(f)$. This is actually equivalent to the derived category of vector spaces, I think is the "simplest one", and still good enough to answer the original question: take A=some nonzero vector space concentrated in degree zero and B=A[1], Hom(A,B)=0 but Hom(A[1],B)=End(A) is not zero