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Let $\mathcal{A}$ be an additive category and $B \to C$ a nonzero map. Are there say "standard" techniques & criteria one should keep in mind when working with additive categories to attack the question if $B$ appears to be direct summand of $C$?
Note that for abelian category "the" standard technique to approach this question is the splitting lemma, unfortunately it is non appliable to additive categories as we have no exact sequences, kernels, cokernels, etc.
But is there a kind of "weakened replacement" of this lemma for additive categories adressing this beeing direct summand issues?

If thats to weak to expect such one, what if our additive category would be additionally triangulated, like bounded derived category of coherent sheaves?
Is there a version of "splitting lemma" for additive + triangulated categories? (...at all, philosophically triangulated property appears to me to be the "next best to abelian" if we want to emulate features of abelian category in additive category). So maybe, we can "emulate" a version of splitting lemma for triangulated category having the case of bounded derived category of coh sheaves in mind.

The background: The question is motivated as suggested by following particular situation appearing in context of bounded derived category of coherent sheaves. More concretely, let $f:X \to Y$ a morphism of schemes and $\mathcal{F}$ a coherent etale sheaf on $Y$ (...we work in following tacitly with etale site). Suppose that $f$ proper, and "nice enough" (eg separated &quasi-finite - so finite by properness) such that there exist a trace map

$$\text{tr}: f_{!}f^*\mathcal{F} \to \mathcal{F}$$

(reference: these notes by Brian Conrad; p54). Moreover we have canonical counit map $c:\mathcal{F} \to f_{*}f^*\mathcal{F}$ and due to assumed finiteness we have $Rf_{!}=Rf_*$ on level of derived functors, so we would obtain in the assoc derived category the composition

$$\mathcal{F} \to Rf_{!}f^*\mathcal{F} \to \mathcal{F}$$

which known to coincide with multiplication by degree of $f$, so non zero map. It looks that $\mathcal{F}$ is rather close to be included in $Rf_{!}f^*\mathcal{F}$ as direct summand. "Rather close" in the sense, as at this stage, if derived category would be abelian category, then the above composition & its properties would be sufficient to have splitting (... also not sure what are the weakest assumptions on $\mathcal{F}$ making the splitting actually be true, eg constructible?)

Does there it this stage a "standard argument" to deduce that $\mathcal{F}$ appears as dirrect summand of $Rf_{!}f^*\mathcal{F} $?
Here I'm also not completely sure - once having written down this composition - which feature of the derived category goes actually in crucial way in to prove the splitting: Do we only need the additivity or is beeing triangulated the crucial ingrediant here? If its only a matter of additiveness, can the argumentation be "standartized" to all additive categories (as kind of "weakened version" replacing splitting lemma) or is it here something specific for those having additionally trianguated structure as it is the can for derived category? (Here we are in the situation that derived cat is only additive, so we cannot invoke the splitting lemma)

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    $\begingroup$ I'm not sure if this is what you're looking for, but one obvious statement is that if you have a map $A\to B$ in a triangulated category, complete it to a triangle $A \to B \to C \to \tau A$. Then $A\to B$ expresses $A$ as a direct summand of $B$ if and only if $C \to \tau A$ is the zero map. In this case we have $B\cong A\oplus C$. $\endgroup$
    – Dave Benson
    Commented Aug 18 at 16:54
  • $\begingroup$ @Dave Benson: On the contrary, I think that's the most "canonical" substitutor of the splitting lemma in triangulated world (...to seek in additive cats alone seems to be to weak as one could expect to have a nice criterion for splitting) one could expect, thanks! Maybe, one could subsequently ponder if the splitting in the example from derived cat actually motivated this question is a consequence of this criterion; this would require that the composition above $F \to f_{!}f^*F \to F$ fits actually in (or is homotopic to) a triangle whose 3th map is zero. Not sure if thats clear? $\endgroup$
    – user267839
    Commented Aug 18 at 17:18
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    $\begingroup$ In a triangulated category, if $A\to B$ has a splitting map (i.e., a map $B\to A$ such that the composition $A\to B\to A$ is $\operatorname{id}_A$), then it follows from @Dave's comment that $A$ is a direct summand of $B$, as then $C\to\tau A$ is equal to the composition $C\to\tau A\to\tau B\to\tau A$, which is zero, since $C\to\tau A\to\tau B$ is zero, since it is the composition of two consecutive maps in a triangle. $\endgroup$
    – Jeremy Rickard
    Commented Aug 19 at 9:07
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    $\begingroup$ In a general additive category, having a splitting map is not enough to ensure being the inclusion of a direct summand. E.g., in the additive category of real vector spaces with dimension not equal to one, an injective map $\mathbb{R}^2\to\mathbb{R}^3$ has a splitting map, but $\mathbb{R}^2$ is not a direct summand of $\mathbb{R}^3$ $\endgroup$
    – Jeremy Rickard
    Commented Aug 19 at 9:10

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