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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?

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?

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

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?

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: In which situations this map splits in sense of that $\mathcal{F} $ becomes direct summand of $Rf_{!}f^*\mathcal{F}$?

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

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