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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?

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.

After trying for a while, I suspect that $F$ faithful is not enough to detect semi-simplicity, so:

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 $D$.

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.

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up vote 3 down vote accepted

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).

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But why is $F$ faithful? – Paul Balmer May 18 '10 at 0:55
Um... forgetful functor? – Victor Protsak May 18 '10 at 2:48
At the derived level it is going to kill the morphisms representing non-trivial extensions though... – Greg Stevenson May 18 '10 at 2:52
@Victor: Take A=kG for a (finite) group G and k a field. In D(kG), the homs from k to suspensions of k are the cohomology groups of G with coefficients in k. When you go to D(k) under the (dangerously named) forgetful functor, all the (positive) cohomology groups go to zero – Paul Balmer May 18 '10 at 3:46
@Paul: You are welcome! I need the following properties: homological dimension 1 (this would be true for any quiver); finitely many indecomposables (this is specific to ADE). Both of these is really needed only in order to avoid infinite dimensional vector spaces in the category $D$. – Victor Ostrik May 18 '10 at 17:14

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