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Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.

Edit 2: In the case of $()^{\ell^n}: \mathbb G_m \to \mathbb G_m$, let us suppose we have a sheaf of the form we want and let the Frobenius have eigenvalues $q\alpha_i$ on degree zero compactly supported cohomology.

We can work out the number $\ell^n$ th powers in $\mathbb F_{q^r}$ and it depends on what power of $\ell$ divides $q^r-1$. If we further assume that $q-1$ is exactly divisible by $\ell$, we get equations of the form $\sum_i \alpha_i^r = \max(\ell^{n-k},\ell)$ for $\ell^{k-1}||r$.

We can solve this using Fourier inversion and it seems the unique solutions give us $\ell^{n-2k}$ copies of $\alpha_i = \zeta_{\ell^k}$ as $\zeta_{\ell^k}$ ranges over all primitive $\ell^k$ roots of unity and $1\leq k \leq n-1$ and $\ell^{n-1}$ copies of $1$.

Unfortunately $\ell^{n-2k}$ is not integral which seems to show that no $\mathbb Q_\ell$ sheaf that is defined by localizing a $\mathbb Z_\ell$ sheaf can work. Is there some other formalism of $\mathbb Q_\ell$ sheaves that can give us something useful?

Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.

Edit 2: In the case of $()^{\ell^n}: \mathbb G_m \to \mathbb G_m$, let us suppose we have a sheaf of the form we want and let the Frobenius have eigenvalues $q\alpha_i$ on degree zero compactly supported cohomology.

We can work out the number $\ell^n$ th powers in $\mathbb F_{q^r}$ and it depends on what power of $\ell$ divides $q^r-1$. If we further assume that $q-1$ is exactly divisible by $\ell$, we get equations of the form $\sum_i \alpha_i^r = \max(\ell^{n-k},\ell)$ for $\ell^{k-1}||r$.

We can solve this using Fourier inversion and it seems the unique solutions give us $\ell^{n-2k}$ copies of $\alpha_i = \zeta_{\ell^k}$ as $\zeta_{\ell^k}$ ranges over all primitive $\ell^k$ roots of unity and $1\leq k \leq n-1$ and $\ell^{n-1}$ copies of $1$.

Unfortunately $\ell^{n-2k}$ is not integral which seems to show that no $\mathbb Q_\ell$ sheaf that is defined by localizing a $\mathbb Z_\ell$ sheaf can work. Is there some other formalism of $\mathbb Q_\ell$ sheaves that can give us something useful?

Alternatively, perhaps we need to assume that the base field is large enough (perhaps large enough to make the galois group of the generic fiber constant?)

Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.

Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.

Edit 2: In the case of $()^{\ell^n}: \mathbb G_m \to \mathbb G_m$, let us suppose we have a sheaf of the form we want and let the Frobenius have eigenvalues $q\alpha_i$ on degree zero compactly supported cohomology.

We can work out the number $\ell^n$ th powers in $\mathbb F_{q^r}$ and it depends on what power of $\ell$ divides $q^r-1$. If we further assume that $q-1$ is exactly divisible by $\ell$, we get equations of the form $\sum_i \alpha_i^r = \max(\ell^{n-k},\ell)$ for $\ell^{k-1}||r$.

We can solve this using Fourier inversion and it seems the unique solutions give us $\ell^{n-2k}$ copies of $\alpha_i = \zeta_{\ell^k}$ as $\zeta_{\ell^k}$ ranges over all primitive $\ell^k$ roots of unity and $1\leq k \leq n-1$ and $\ell^{n-1}$ copies of $1$.

Unfortunately $\ell^{n-2k}$ is not integral which seems to show that no $\mathbb Q_\ell$ sheaf that is defined by localizing a $\mathbb Z_\ell$ sheaf can work. Is there some other formalism of $\mathbb Q_\ell$ sheaves that can give us something useful?

Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.

Edit 2: Looking at the example of $(.)^\ell: \mathbb G_m \to \mathbb G_m$, the number of $\mathbb F_{q^r}$ points depends on whether or not $\ell | q^r-1$ (it's either $(q^r-1)$ or $(q^r-1)\ell$ after multiplying by the degree). If we suppose that $q$ is a generator for $(\mathbb Z/\ell)^\times$, then it seems that on the degree zero compactly supported cohomology, we want the eigenvalues to be $q\zeta_{\ell-1}^{k}$ for $1 \leq k\leq \ell$ and similarly for degree $0$.

Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?

If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.

Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.