### Definitions


Let $\mathcal{T}$ be a triangulated category with translation functor $\Sigma$.


* **Generator**: An object $T$ of $\mathcal{T}$ is a generator of $\mathcal{T}$ if $\mathcal{T}(\Sigma^n T,A)=0$ for all $n$ implies $A=0$.
* **$T$-ghost**: A map $f:A\to B$ in $\mathcal{T}$ is called $T$-ghost if the reduced map
$$
\textsf{Hom}\_{\mathcal{T}}(\Sigma^n T,f) : \textsf{Hom}\_{\mathcal{T}}(\Sigma^n T,A) \to \textsf{Hom}\_{\mathcal{T}}(\Sigma^n T,B)
$$
is zero for all $n\in\mathbb{Z}$.
* **$T$-ghost ideal**: We denote by $\textsf{Gh}\_{T}(A,B)$ the collection of all $T$-ghost maps between $A$ and $B$ and
$$
\textsf{Gh}\_{T}(\mathcal{T}):=\bigcup\_{A,B\in\mathcal{T}}\textsf{Gh}\_T (A,B)
$$
Clearly $\textsf{Gh}\_{T}(\mathcal{T})$ is an ideal of $\mathcal{T}$, called the $T$-ghost ideal of $\mathcal{T}$.
* **Jacobson radical**: For any objects $A, B\in \mathcal{T}$, the subgroup $\textsf{Rad}(A,B)$ of $\mathcal{T}(A,B)$ consists of all maps $f:A\to B$ such that $1\_A - gf: A\to A$ is invertible, for any map $g: B\to A$. Now we define $\textsf{Rad}(\mathcal{T})$ as the collection of all subgroups $\textsf{Rad}(A,B)$:
$$
\textsf{Rad}(\mathcal{T})=\bigcup\_{A,B\in\mathcal{T}}\textsf{Rad}(A,B)
$$




---


Could you please help me with the Remark 2.9 in [this article](https://www.math.uni-bielefeld.de/%7Esek/2008/ghosts.pdf)? Thank you very much!



> 
> Let $\mathcal{T}$ be a triangulated category with translation functor $\Sigma$, let $T$ be an object of $\mathcal{T}$. Then the following are equivalent:
> 
> 
> (i) $T$ is a generator of $\mathcal{T}$.
> 
> 
> (ii) The ideal of $T$-ghost maps is contained in the Jacobson radical $\textsf{Rad}(\mathcal{T})$ of $\mathcal{T}$.
> 
> 
> 


*Proof*. (i) $\implies$ (ii). Let $f: A\to B$ be a $T$-ghost map. If $A=0$, then $gf = 0$ for all $g: B\to 0$ ($\color{red}{\text{right?}}$), so $1\_A - gf = 1\_A$ is invertible. If $A\ne 0$, then there is a morphism $l: \Sigma^n T \to A$ such that $f\circ l = 0$, hence $(1\_A - gf)l = l$. Now how to show that $1\_A - gf$ is invertible?


(ii) $\implies$ (i).



$T$ is a generator of $\mathcal{T}$ iff. the ideal of $T$-ghost maps is contained in the Jacobson radical of $\mathcal{T}$
(i) $\implies$ (ii): If $f \colon A \to B$ is a $T$-ghost and $g \colon B \to A$ is arbitrary, then $gf$ is a $T$-ghost and therefore $1\_A - gf$ induces the identity map $\mathsf{Hom}(\Sigma^n T,A) \to \mathsf{Hom}(\Sigma^n T,A)$ for all $n$. If $C$ is the cone of $1\_A -gf$ then the long exact sequence for Hom shows that $\mathsf{Hom}(\Sigma^n T, C)=0$ for all $n$, so $C=0$ since $T$ is a generator, so $1\_A-gf$ is invertible.


(ii) $\implies$ (i): If $A$ is such that $\mathsf{Hom}(\Sigma^n T,A)=0$ for all $n$ then the identity map $1\_A$ is $T$-ghost. But if the identity map is in the Jacobson radical then the zero map from $A$ to $A$ is invertible, which implies $A=0$.


