I am reading the book Elements of the Representation Theory of Associative Algebras Volume 1 written by Ibrahim Assen, Daniel Stimson and Andrzej Skowronski and the paper on the radical of a category written by G.M. Kelly.
Let $R$ be a finite dimensional $K$-algebra and we only consider $\mathrm{mod}~R$ whose objects are the finitely generated right $R-$modules.
In the above book and paper, they define the radical of $\mathrm{Hom}_R(A,B)$:
$$\mathrm{rad}_R(A,B)=\{f\in \mathrm{Hom}_R(A,B)~|~1_A-gf ~~\text{is invertible}, \forall g\in \mathrm{Hom}_R(B,A)\}.$$
While I read the book Representation Theory of Artin Algebras, in this book, the authors define the radical of $\mathrm{Hom}_R(A,B)$:
$\mathrm{rad}_R(A,B)=\{f\in \mathrm{Hom}_R(A,B)~|~hfg~\text{is not an isomorphism}, $
$~~~~~~~~~~~~~~~~~~~~~~~~~~\text{for any}~g\in \mathrm{Hom}_R(X,A) ~\text{and any}~h\in$ $ ~~~~~~~~~~~~~~~~~~~~~~~~\mathrm{Hom}_R(B,X) ~\text{with }~X~\text{is indecomposable in} ~\mathrm{mod}R\}$
For convenience, we denote the first definition and the second definition by $R_1$ and $R_2$, respectively.
I want to get the equivalence of two definitions of $\mathrm{rad}_R(A,B)$, i.e. $R_1=R_2.$
Here are my attempts:
If we choose $f\in R_1$, we want to prove that $hfg$ is not an isomorphism.
If not, there exits $\nu\in \mathrm{Hom}_R(X,X)$ such that $(hfg)\nu=1_X=\nu (hfg)$.
Then we have $g\nu h\in \mathrm{Hom}_R(B,A)$ and, by the definition of $R_1$, $1_A-(g\nu h)f$ is invertible.
Thus we have $s\in \mathrm{Hom}_R(X,X)$ such that $s(1_A-(g\nu h)f)$ $=1_A.$By the right multiplication of $g$, we have $s(1_A-(g\nu h)f)g=g.$
Since $\nu (hfg)=1_X$,we get $g=0$. Then $hfg=0$ is not an isomorphism and it is a contradiction. Thus we prove $R_1\subset R_2.$
But I have no idea to prove $R_2\subset R_1$.
Any help and references are greatly appreciated.
Thanks!