# Why $R^q(\Gamma \circ \eta_{*}) (\Bbb G_{m, \eta}) = H^q(\eta_{ét}, \Bbb G_{m, \eta})$?

Let $X$ be a smooth, projective and connected curve over an algebraically closed field, and let $\eta \rightarrow X$ be its generic point (we also call the inclusion as $\eta$). I want to understand the proof of the computation of the étale cohomology of $\Bbb G_{m, \eta}$ on $X$. Reading SGA 4.5, by P. Deligne, I got stacked at the lemma 3.3 of the section "Cohomolgie des courbes" of . It says the following:

$H^q(X_{ét}, \eta_* \Bbb G_{m, \eta})=0$ for $q >0$.

In the proof, they consider the exact sequence arising from the spectral sequence given by the functors $\eta_*: Ab(\eta_{ét}) \rightarrow Ab(X_{ét})$ and $\Gamma(X_{ét}, -): Ab(X_{ét}) \rightarrow Ab$ (I will denote this functor as $\Gamma$), where Ab(-) are the sheaves of abelian groups. More precisely, we obtain the exact sequence $$0 \rightarrow R^1\Gamma(\eta_* \Bbb G_{m, \eta}) \rightarrow R^1(\Gamma \circ \eta_*)(\Bbb G_{m, \eta}) \rightarrow \Gamma \circ R^1(\eta_* \Bbb G_{m, \eta}) \rightarrow \cdots.$$

Since the $R^q(\eta_* \Bbb G_{m, \eta})$ are zero, we get isomorphisms $H^q(X_{ét}, \eta_* \Bbb G_{m, \eta}) \cong R^q(\Gamma \circ \eta_*)(\Bbb G_{m, \eta})$.

So the question is, why $R^q(\Gamma \circ \eta_{*}) (\Bbb G_{m, \eta}) = H^q(\eta_{ét}, \Bbb G_{m, \eta})$? Once this equality is stablished, a previous result concludes the proof.

We do this by steps using the same thing: for any $f:X \rightarrow Y$ we have that $\Gamma(X_{ét},-) = \Gamma(Y_{ét}, f_*-)$. Let $\pi: X \rightarrow Spec(k)$ be the structure morphism. In the formulae, we denote $\Gamma_k := \Gamma((Spec k)_{ét}, -)$ and $\Gamma := \Gamma(X_{\acute{e}t},-)$.
Step 1: Here, $\Gamma (X_{ét}, -) = \Gamma ((Spec \: k)_{ét}, \pi_*-)$ imples $$R^q(\Gamma \circ \eta_*)(\Bbb G_{m, \eta}) = R^q(\Gamma_k \circ \pi_* \circ \eta_*)(\mathbb{G}_{m, \eta}).$$ Step 2: Here, $\Gamma(\eta_{ét}, -) = \Gamma (X_{ét}, \eta_*-)$ implies $$R^q(\Gamma_k \circ (\pi \circ \eta)_* \Bbb G_{m, \eta}) = R^q\Gamma (\eta_{ét}, \Bbb G_{m, \eta}).$$
and we are done, since the last term is by definition $H^q(\eta_{\acute{e}t}, \Bbb G_{m, \eta})$. Note that in the rhs, $\Gamma$ is not the same as in the previous equality, sorry for the notation!