Recall that in a triangulated category, all monomorphisms split (have a retraction). Let $F:C\to D$ be an exact functor between triangulated categories. It is an easy exercise to see that if $F$ is faithful then it detects monomorphisms: If $F(f)$ is a monomorphism then so is $f$. Same with epimorphisms of course, and with isomorphisms. But what about semi-simplicity in the following sense? 



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> **Definition**: Let us say that a morphism $f$ is *semi-simple* if there exists $g$ in the opposite direction such that $f=fgf$. (BTW, what is the right name for this?) Assuming $C$ idempotent-complete, $f$ is semi-simple $\iff$ $f$ is a composition of a split epimorphism, an isomorphism and a split monomorphism $\iff$ the exact triangle over $f$ is a sum of trivial triangles. 
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After trying for a while, I suspect that $F$ faithful is not enough to detect semi-simplicity, so:



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> **Problem**: Find an exact faithful functor $F:C\to D$ between idempotent-complete triangulated categories and a morphism $f$ in $C$ which is not semi-simple but such that $F(f)$ is semi-simple in $D$.
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Of course, it might also be true that faithfulness detects semi-simplicity. A proof of that would certainly count as an answer to the above problem!


What would make me really happy would be to have $F=(\ F:C\to C\ ,\ \mu:F^2\to F\ ,\ \eta:Id\to F\ )$ an exact faithful *monad* (a.k.a. triple) on the category $C=D$. In that case, $F$ faithful forces the unit $\eta$ to be objectwise a split monomorphism. If $\eta$ has moreover a natural retraction then $F$ detects semi-simplicity (easy) but this naturality of the retraction definitely fails for general monads.



Do exact faithful functors between triangulated categories detect semi-simplicity?
Let $A$ be a non semi-simple algebra and $C$ be the derived category of finite dimensional
$A-$modules. Let $D$ be the derived category of finite dimensional vector spaces and let
$F$ be the forgetful functor which maps $A-$module to its underlying vector space. 
Then $F(f)$ is semi-simple for any $f$ (since any morphism in $D$ is semisimple). 


UPDATE: as explained below this is not an answer since $F$ is not faithful. Here is an actual example: take for $C$ derived category of representations of an ADE quiver; let $D$ be as above and
let $F=RHom(T,?)$ where $T$ is a direct sum of all indecomposable representations of the quiver
(there are just finitely many of them). Then $F(f)$ is semi-simple for any $f$ as before; to check
that $F$ is faithful use the fact that any object of $C$ is isomorphic to direct sum of shifted
indecomposable representations of the quiver (since the category of representations of the quiver
has homological dimension 1).


