Splitting of composition of trace and counit in derived setting

Question

Let $X,Y$ be varieties (separated of finite type schemes) over base field $k$, $\mathcal{F}$ be constructible sheaf on $Y_{\mathrm{et}}$ and assume that we have a finite morphism $f: X \to Y$, which guarantees (see e.g. these notes on p54) there exist a well-defined "trace map"

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

As finite morphisms are proper we have $f_*=f_{!}$ and so we can precompose with counit map obtaining the composition which transfers into derived world - we work here with derived category $D_c^b(X)$ of constructible sheaves on $X_{et}$ (...or, if we additionally require $\mathcal{F}$ to be $l$-adic sheaf (for $\ell$ prime in $k$) we shall also reason on derived cat $D_c^b(X, \ell)$ of constructible $\ell$-adic sheaves):

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

Question: Which assumptions on $\mathcal{F}$ guarantee that this map actaully splits in sense of that $\mathcal{F} $ becomes direct summand of $Rf_{!}f^*\mathcal{F}$ in derived category?

It is well known that this composition coincides with multiplication by degree of $f$, and the I'm wondering when we actually have that $\mathcal{F}$ appears as direct summand of middle object.
An idea: can we recognize this composition as part of an valid exact triangle with one zero map, compare with this question?

Answer 1

Jeremy Rickard's argument in the comments of your other question gives the solution. If $f \colon A \to B$ and $g \colon B \to A$ have $g \circ f$ an isomorphism $A \to A$, then there exists $g'\colon B \to A$ such that $g' \circ f $ is $id_A$. Indeed, we can just take $g'=(g \circ f)^{-1} \circ g$.

Now form the distinguished triangle $C[-1] \to A \to B \to C$ completing $f$. Then $C[-1]\to A$ composed with $f$ is zero, so composed with $g \circ f$ is zero, but $g\circ f$ is the identity, so $C[-1]\to A$ must be $0$. Now since one morphism of the distinguished triangle is isomorphic to the distinguished triangle $ A\to A\oplus C \to C $, the whole triangle must be isomorphic to the distinguished triangle $A \to A\oplus C \to C$, giving the desired splitting.