Is it always possible to compute the Betti numbers of a nice space with a well-chosen Lefschetz zeta function?

Question

Let $X$ be a smooth projective variety. If I've understood correctly, the Weil conjectures imply that it is possible to compute the Betti numbers of $X(\mathbb{C})$ by computing the local zeta function associated to $X(\overline{\mathbb{F}_p})$ for $p$ a prime of good reduction. This is because one can identify the factors coming from different $\ell$-adic cohomology groups by the absolute value of their roots (the "Riemann hypothesis," by analogy with the case of curves).

Unfortunately, this doesn't seem to be true for the analogous situation with compact triangulable spaces $Y$ and the Lefschetz zeta function $\zeta_f$, at least not for an arbitrary choice of function $f : Y \to Y$. For example, if $f$ is homotopic to the identity, then $\zeta_f$ can only see the Euler characteristic of $Y$. Even if $f$ acts "generically" (i.e. none of the factors of $\zeta_f$ cancel), there doesn't seem to be a way to distinguish which factors are associated to which degree. (Of course, I would love if I were wrong about this.)

Question 1: When is there an analogue of the geometric Frobenius for compact triangulable spaces $Y$? By this I mean a more-or-less canonical function $f : Y \to Y$ such that some analogue of the Riemann hypothesis holds for $\zeta_f$.

Question 2: Regardless of the answer to Question 1, is it always possible to choose $f$ such that $\zeta_f$ can tell you which of its factors are associated to which homology groups?

(Side question: is it true that given any $f : Y \to Y$ there is always some $f'$ homotopic to $f$ which has finitely many fixed points?)

Answer 1

Having resolved my ignorance concerning surface groups I can now answer question 1 negatively (or at least some formulation thereof). It is impossible if $Y$ is an oriented surface of genus at least $2$.

Suppose that $f: Y \to Y$ is a self map of the surface such that the eigenvalues of $f^*$ acting on each $H^i(Y)$ are all nonzero (otherwise we can't "detect" the betti numbers), and such that $H^i(Y)$ and $H^j(Y)$ do not have eigenvalues of common magnitude for $i \neq j$. Then in particular $f^*$ acts on $H^2(Y)$ nontrivially, say by multiplication by some integer $d$. This integer cannot be $\pm 1$ since then $H^0(Y)$ and $H^2(Y)$ would contain eigenvectors with eigenvalues of equal magnitude.

Consider the subgroup $H = f_*(\pi_1(Y))$ inside $G = \pi_1(Y)$. If this had infinite index, then $f$ would lift to a map to some infinite covering of $Y$, so it would induce a trivial map of $H^2$. So $H$ has finite index in $G$. Let $X \to Y$ be the corresponding covering space. Then $\pi_1(X)$ is a quotient of $\pi_1(Y)$, hence its abelianization has rank $\leq 2g$ where $g$ is the genus of $Y$. This implies that $X$ is a closed surface of genus at most $g$. But its Euler characteristic is precisely $[G:H]$ times the Euler characteristic of $Y$, so $X = Y$. Thus $f$ induces a surjection on $\pi_1(Y)$. By the post cited above, $f$ actually induces an isomorphism on $\pi_1(Y)$, so it is a homotopy equivalence. In particular, $d = \pm 1$, contrary to assumption.

After writing this it occurs to me that you might object to me ruling out the case $d = -1$... At any rate, this shows that the eigenvalues can't ever look like they do in the case of the Riemann hypothesis, with magnitude $q^{i/2}$ on $H^i$ for some $q>1$.