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My question came up while reading this article by Nicholas Katz, specifically lemma 4.2. I don't think it's necessary to read the article to answer the question, but I'm including it anyways for reference.

Suppose we have a smooth, affine, geometrically integral curve $U_0/\mathbb{F}_{q}$, and let $U/\overline{\mathbb{F}}_{q}$ be it's base change. Then we have the following exact sequence:

$$1 \rightarrow \pi_{1}(U) \rightarrow \pi_{1}(U_{0}) \rightarrow Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \rightarrow 1$$

One has the frobenius $Frob_{q}\in Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q})$, which acts on $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ by transport of structure.

In this proof, we know that $Frob_{q}$ does not have $1$ as an eigenvalue, so that it operates without fixed points. To prove his result he derives a contradiction by constructing a nontrivial continuous additive homomorphism $f:\pi(U_{0})\rightarrow \overline{\mathbb{Q}}_{\ell}$, and claims that by restriction to $\pi_{1}(U)$ we get a fixed point in $H^{1}(U,\overline{\mathbb{Q}}_{\ell})$.

I'm aware of the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$, and as far as I can tell the proof of the claim would go like this. Use the homomorphism $Gal(\overline{\mathbb{F}}_{q}/\mathbb{F}_{q}) \hookrightarrow Aut(\pi_{1}(U))/Inn(\pi_{1}(U))$ to get an action of $Frob_{q}$ on $\pi_{1}(U)$. For $\phi\in Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})$ define $(Frob_{q}\cdot\phi)(a) = \phi(a^{Frob_{q}})$. Because our $f$ mentioned above is restricted from $\pi_{1}(U_{0})$ where it also factors through $\pi_{1}(U_{0})^{ab}$, $f$ is indeed invariant under this action.

Essentially what I don't see is why the isomorphism $Hom_{cont}(\pi_{1}(U),\overline{\mathbb{Q}}_{\ell})\simeq H^{1}(U,\overline{\mathbb{Q}}_{\ell})$ commutes with the action of $Frob_{q}$, which seems to be necessary. I only know the above isomorphism through a chain of fairly non-constructive isomorphisms-- something along the lines of what's mentioned in 11.3 of Milne's notes. However, it seems unlikely that something like this wouldn't turn out to be true. Can anyone explain why?

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  • $\begingroup$ The clue I have is: Look at Prop 1.3.4 in Deligne's Weil II $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2014 at 13:43
  • $\begingroup$ Which, in turn, can be found here: math.harvard.edu/~gaitsgde/grad_2009/Weil-II.pdf $\endgroup$
    – Igor Rivin
    Commented Mar 25, 2014 at 13:52
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    $\begingroup$ I don't know what you mean by constructive, but all the isomorphisms seem constructive to me, and are perfectly functorial. THe point is that even if you choose some object from an equivalence class and apply some construction to it, checking that equivalent inputs give you equivalent outputs also shows that the construction commutes with everything that commutes with your proof. $\endgroup$
    – Will Sawin
    Commented Apr 9, 2014 at 15:04
  • $\begingroup$ im speaking more to the fact that the frobenius already has an action on H^{1}(U), and here i could define another. it's not obvious to me that they are the same. $\endgroup$
    – martyall
    Commented Apr 11, 2014 at 2:56

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