Pushforward of constant sheaf on the generic point. Let $X$ be an integral scheme, and $g: \eta \to X$ the inclusion of its generic point. If $M$ is an abelian group and $M_\eta$ is a constant sheaf on $\eta$ taking value $M$, then is it true that $g_* M_\eta = M_X$? 
My thought is yes, as if $U \subset X$ is open, then $g_* M_\eta (U) = M_\eta( g^{-1}(U)) = M$, as every nonempty open set will contain the generic point. However it seems that one may need that $X$ is normal (or some other hypothesis) for this to be true. For reference, this is Exercise 3.7 (Chap II) in Milne's Etale Cohomology.
 A: We need to show
$$M_X\rightarrow g_*g^*M_X=g_*M_\eta\tag{$\dagger$}$$
is an isomorphism. Fix a geometric point $x$ of $X$ and let $X_{(x)}$ denote the (spectrum of the) strict henselization of $X$ at $x$; $X_{(x)}$ is a normal domain as $X$ is normal. The map on stalks is
$$M\rightarrow\Gamma(\eta\times_X X_{(x)},M_\eta).$$
$\eta_x:=\eta\times_X X_{(x)}$ is the spectrum of the field of fractions of the strict henselization of $X$ at $x$, which is separable algebraic over $k(\eta)$.
Letting $I_x:=\operatorname{Gal}(k(\overline\eta)/k(\eta_x))$ ($k(\overline\eta)$ denotes the separable closure of $k(\eta_x)$, which is the same as the separable closure of $k(\eta)$),
$$\Gamma(\eta\times_X X_{(x)},M_\eta)=M^{I_x}:=\{m\in M:gm=m\forall g\in I_x\}$$
where here we have identified $M$ with the stalk of $M_\eta$ at a geometric point centered on $\eta$. So asking that the map ($\dagger$) be an isomorphism at the geometric point $x$ is the same as requiring that $I_x$ act trivially on $M$ considered as $\operatorname{Gal}(k(\overline\eta)/k(\eta))$-module. In particular, this condition is satisfied by the constant sheaf, which corresponds to $M$ with trivial action of Galois.
(I am implicitly using here the description of étale morphism with normal integral target in EGA $\text{IV}_4$ 18.10.7.)
