Background: For a smooth proper variety $X$ over an algebraically closed field $k$, we have the etale cohomology groups $H^i(X,\mathbb{Q}_{\ell})$ for $\ell \not= p$. We can use the Kummer exact sequence to show that $H^1(X,\mathbb{Q}_\ell(1))$ is isomorphic to $V_\ell \mathrm{Pic}^0(X)$, where $\mathrm{Pic}^0(X)$ is the Picard variety of $X$, and $V_\ell(-)$ indicates taking the inverse limit of $\ell^n$-torsion points as $n \rightarrow \infty$ (i.e., the Tate module), and tensoring with $\mathbb{Q}$. Moreover, the Kummer exact sequence also shows that for a morphism of smooth proper varieties $f: X \rightarrow Y$, the pullback map $f^*: H^1(Y,\mathbb{Q}_\ell(1)) \rightarrow H^1(X,\mathbb{Q}_\ell(1))$ agrees with the map induced by pullback of line bundles $(L \mapsto f^*L): V_\ell\mathrm{Pic}^0(Y) \longrightarrow V_\ell \mathrm{Pic}^0(X)$.

Suppose moreover that $f: X \rightarrow Y$ is a map of smooth proper varieties of the same dimension $d$. Then Poincare duality shows that $H^1(X,\mathbb{Q}_{\ell}(1))$ is dual to $H^{2d-1}(X,\mathbb{Q}_{\ell}(d-1))$, and we can take the dual to the pullback map $H^{2d-1}(Y,\mathbb{Q}_\ell(d-1)) \rightarrow H^{2d-1}(X,\mathbb{Q}_{\ell}(d-1))$ to get a map $$f_*: H^1(X,\mathbb{Q}_\ell(1)) \rightarrow H^1(Y,\mathbb{Q}_\ell(1)).$$

Here's my question: How do we describe this map in terms of line bundles? Does it agree with the map $V_\ell \mathrm{Pic}^0(X) \rightarrow V_\ell \mathrm{Pic}^0(Y)$ induced by proper pushforward of divisors (i.e., represent each line bundle by a divisor and take its proper pushforward)? This seems extremely likely to me, but I'm not sure how to prove it in general. Notice that this qustion is not the same as the compatibility of cycle class maps with pushforward, since the cycle class of a divisor lives in $H^2$, not $H^1$.

`$f_* f^*F \rightarrow F$`

, and I think I can show that when`$F = \mu_n$`

, this is given by the norm function as you say. When $f$ has fibers of dimension $\geq 1$, the map must be 0. The last case is when $f$ is generically finite, which one does by shrinking on $X$ and $Y$. $\endgroup$ – Peter Mar 19 '13 at 19:05