This should a be basic enough question, but I’m a little confused.

In proving that $H^*(X,\mathbf{Q}_{\ell})$ is functorial (in the sense of Weil cohomology theories: see axiom D2 here) as $X$ ranges over smooth projective varieties over an algebraic closure $k$ of a finite field, one argues proving that for any map $f : X\to Y$ of smooth projective varieties over $k$, we have that the natural map

$$\alpha_f : f^{-1}\mathbf{Q}_{\ell}\to \mathbf{Q}_{\ell}$$

in $D_{\rm cons}(X,\mathbf{Q}_{\ell})$, is an isomorphism.

Here the inverse image functor is constructed as $(\varprojlim\ f^{-1}_{\rm ét}\mathbf{Z}/\ell^n\mathbf{Z})[1/\ell]$ so to speak, where $f_{\rm ét}^{-1}$ is the étale inverse image on abelian sheaves.

Why is it so important to know that $\alpha_f$ is an isomorphism? One always has a map $\alpha_f$, and could define the effect of $f$ on cohomology, $f^* : H^*(Y,\mathbf{Q}_{\ell})\to H^*(X,\mathbf{Q}_{\ell})$, to be the composition:

$$H^*(Y,\mathbf{Q}_{\ell})\to H^*(X, f^{-1}\mathbf{Q}_{\ell})\xrightarrow{\alpha_f} H^*(X,\mathbf{Q}_{\ell})$$

where the first map always exists by functoriality of the étale site, without knowing that $\alpha_f$ is an isomorphism.

Where is the fact that $\alpha_f$ is an isomorphism, so crucially used? Is it used in showing that $f^*$ has good properties: namely, it induces a map of graded commutative $\mathbf{Q}_{\ell}$-algebras? Or perhaps on an even more basic level, to show that for a composition $f\circ g$, we have $(f\circ g)^* = g^*\circ f^*$?


As you say, one can talk about functoriality quite generally and obtain maps of the form $$f^* : H^n(Y, F) \to H^n(X, f^{-1} F),$$ for any ($\ell$-adic, if you want) sheaf $F$. This will be appropriately functorial. And then, if you have amorphism of sheaves on $X$, $\alpha: f^{-1} F \to G$, then you can append, as you say, the induced map on cohomology, $$H^n(X, f^{-1} F) \to H^n(X, G).$$ This does work when the map $\alpha$ is not an isomorphism.

However, the fact that $\alpha: f^{-1} \mathbf Q_\ell \to \mathbf Q_\ell$ is an isomorphism is, in my opinion, a very useful fact (and much more basic than the definition of cohomology groups, say). In fact, to simplify the statement, let me just state it for sheaves of abelian groups on the small Zariski site (the statement for $\ell$-adic sheaves on the étale site is similar though): then $f^{-1}$ is the unique colimit-preserving functor (between the categories of sheaves on $Y$ and $X$) which satisfies $$f^{-1}(\mathbf Z_U) = \mathbf Z_{f^{-1}(U)}.$$ Here $\mathbf Z_?$ denotes the representable sheaf associated to an open $U \subset X$.

A different perspective on the question is this: the cohomology group $H^n(X, \mathbf Z/\ell)$ (and similarly with $\ell$-adic coefficients) is the group of homomorphisms $Hom_{D(Shv(X))}(\mathbf Z/\ell, \mathbf Z/\ell[n])$ in the derived category of (étale) sheaves on $X$. The map $f^*$ in question identifies, under this identification, with the map which sends a map $\mathbf Z/\ell \stackrel r \to \mathbf Z/\ell[n]$ to the induced map (given by functoriality) $f^{-1} \mathbf Z/\ell \stackrel{f^{-1}(r)} \to f^{-1} \mathbf Z/\ell$. (It would make sense to write $f^* \mathbf Z/\ell$ instead, I am just sticking to the convention used in the question). So, once again it comes down to identifying $f^{-1} \mathbf Z/\ell$ with $\mathbf Z/\ell$. This identification is completely natural. The algebra structure (in the axiom D2 you mention) on $H^*(X)$ arises, in the above interpretation, from the composition of morphisms in the derived category, so no additional facts are needed to show that $f^*$ is a map of algebras.

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  • $\begingroup$ I’d like to ask your thoughts about this. True, you’re right that one can append the induced map of any map of sheaves $\alpha$, but for the purpose of checking that $H^n(X,F)$ satisfies axiom D2 of a Weil cohomology, one should make sure $F=G$ and $\alpha$ is natural. In fact, not only this. $F$ should be a sheaf of algebras over a field $K$ of characteristic zero, and for $n=0$ the resulting map should be a map of $K$-algebras. Maybe this is the point? The isomorphism $f^{-1}\mathbf{Q}_{\ell}\simeq\mathbf{Q}_{\ell}$ is obviously natural $\endgroup$ – user134132 Jan 3 '19 at 14:59
  • $\begingroup$ and makes the map on $H^0$ into a $\mathbf{Q}_{\ell}$-algebra map, in particular $\mathbf{Q}_{\ell}$-linear. $\endgroup$ – user134132 Jan 3 '19 at 15:00
  • $\begingroup$ I have elaborated further. Yes, $\alpha$ is completely natural (the adjoint of the natural map $\mathbf Q_\ell \to f_* \mathbf Q_\ell$), and the existence of this isomorphism $\alpha$ yields that $f^*$ is a map of algebras. $\endgroup$ – Jakob Jan 3 '19 at 22:36
  • $\begingroup$ Upon further thought, although this should be basic enough, it’s not actually clear what the map $\mathbf{Q}_{\ell}\to f_*\mathbf{Q}_{\ell}$ should be. My guess is that, since over each $U$ étale over $Y$, $\Gamma(U_{ét},f_*\mathbf{Q}_{\ell}) = \Gamma(f^{-1}(U)_{ét},\mathbf{Q}_{\ell})$, we can now use the natural isomorphism you wrote: $(\mathbf{Q}_{\ell})_{f^{-1}(U)} = f^{-1}((\mathbf{Q}_{\ell})_U)$, and the map is the map induced by functoriality $\Gamma(U,\mathbf{Q}_{\ell})\to \Gamma(f^{-1}U,f^{-1}\mathbf{Q}_{\ell})$, $\endgroup$ – user134132 Jan 4 '19 at 19:23
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    $\begingroup$ The map $\mathbf Q_\ell \to f_* \mathbf Q_\ell$ is deduced from a similar map $\mathbf Z \to f_* \mathbf Z$. This is defined on sections over $U \subset X$ in the natural way: $\mathbf Z(U)$ consists of the $\mathbf Z$-valued functions on $U$ which are constant on each connected component, similarly for $(f_* \mathbf)(U)$, which are functions on $f^{-1}(U)$ constant on each component. The map is given by sending a function $g: U \to \mathbf Z$ to $g \circ f: f^{-1}(U) \to \mathbf Z$. That's it. $\endgroup$ – Jakob Jan 5 '19 at 20:23

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