Lemma 59.69.2. Let $\pi : X \to Y$ be a nonconstant morphism of smooth projective curves over an algebraically closed field $k$ and let $n \geq 1$ be invertible in $k$. The map

is given by multiplication by the degree of $\pi $.

Lemma 59.69.2. Let $\pi : X \to Y$ be a nonconstant morphism of smooth projective curves over an algebraically closed field $k$ and let $n \geq 1$ be invertible in $k$. The map

\[ \pi ^* : H^2_{\acute{e}tale}(Y, \mu _ n) \longrightarrow H^2_{\acute{e}tale}(X, \mu _ n) \]

is given by multiplication by the degree of $\pi $.

**Proof.**
Observe that the statement makes sense as we have identified both cohomology groups $ H^2_{\acute{e}tale}(Y, \mu _ n)$ and $H^2_{\acute{e}tale}(X, \mu _ n)$ with $\mathbf{Z}/n\mathbf{Z}$ in Lemma 59.69.1. In fact, if $\mathcal{L}$ is a line bundle of degree $1$ on $Y$ with class $[\mathcal{L}] \in H^1_{\acute{e}tale}(Y, \mathbf{G}_ m)$, then the coboundary of $[\mathcal{L}]$ is the generator of $H^2_{\acute{e}tale}(Y, \mu _ n)$. Here the coboundary is the coboundary of the long exact sequence of cohomology associated to the Kummer sequence. Thus the result of the lemma follows from the fact that the degree of the line bundle $\pi ^*\mathcal{L}$ on $X$ is $\deg (\pi )$. Some details omitted.
$\square$

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