Why are morphisms between non-isomorphic indecomposable modules always radical? I am trying to follow the construction of the Auslander-Reiten quiver of the Kronecker algebra, given by Barot in his book Introduction to the Representation Theory of Algebras.
There, he defines a morphism $f: M \to N$ between $A$-modules $M$, $N$ to be radical, if for every morphism $g: N \to M$ it holds true that $\xi = 1_M - g \circ f$ is an isomorphism.
Now, in the proof of lemma 6.2, he states:

[...] Since morphisms between non-isomorphic indecomposables are always radical [...]

I am trying to figure out, why this is the case. I found out that $\ker(\xi)$ cannot be $M$, as this would mean that $g$ is a retraction and hence $N \cong M \oplus \ker(g)$ which yields $M=0$ or $M \cong N$, both of which can not hold by assumption.
However, I think that I mixed up the terms irreducible module and indecomposable module and hence I am not sure how to proceed from here (if it is even possible). 
 A: Barot is using Example 3.23 of the same book,  which is the following:

By a spectroid he means a $k$-linear category $\mathcal{C}$  such that


*

*$\mathcal{C}(x,y)$ is finite dimensional;

*$\text{End}_{\mathcal{C}}(x,x)$ is local for each $x\in\mathcal{C}$;

*objects are pairwise non-isomorphic.


For your ends, he  is considering the category whose objects are indecomposable $A$-modules.
Now, the proof of this example is not given, or at least I couldn't find it, in the book. However, the same result is in appendix A3 of Assem-Simson-Skowronski:
and below is the proof of (b):
If you  don't have access to this book, the result (I.1.3) is the following result.

Lemma
  Let $A$ be a $k$-algebra, then the following are equivalent for $a\in A$:
  
  
*
  
*$a\in\text{rad}\,A$;
  
*$a$ is in the intersection of all left maximal ideals of $A$;
  
*for any $b\in A$, the element $1-ab$ has a two sided inverse;
  
*for any $b\in A$, the element $1-ab$ has a right inverse;
  
*for any $b\in A$, the element $1-ba$ has a two sided inverse;
  
*for any $b\in A$, the element $1-ba$ has a left inverse.
  

