The fiber of the sheaf of invariants Let us suppose the the group $G:=\mathbb{Z}/2\mathbb{Z}=(1,i)$ freely act on a smooth projective variety/$k$ $X$ and denote by $Y$ the G.I.T. quotient $X/G$. Let $\pi:X\longrightarrow Y$ the quotinet map. Now take a $G$-linearised coherent sheaf $(\mathcal{F}, \lambda)$, one can construct the sheaf of invariants of $\mathcal{F}$,the sheaf $\pi_\ast(\mathcal{F})^G$, that  is a sheaf on $Y$. Then given $\overline{\alpha}=\pi(\alpha)$ a point of $Y$ it is possible to consider the fiber of  $\pi_*(\mathcal{F})^G$ at $\overline{\alpha}$, $\pi_\ast(\mathcal{F})^G(\overline{\alpha})$. On the other and one can consider the fiber of $\pi_\ast(\mathcal{F})(\overline\alpha)$ that is (I think) isomorphic to the direct sum
$$\mathcal{F}(\alpha)\oplus\mathcal F(i\cdot\alpha)\simeq \mathcal F\otimes(k(\alpha)\oplus k(i\cdot\alpha))$$
Now this vector space admit a $G$-action induced by $\lambda$, given by $\lambda\otimes\sigma$ where sigma is the action on $k(\alpha)\oplus k(i\cdot\alpha)$ given by permutation.
My question is: the fiber of the sheaf of invariants is isomorphic to the invariant subspace of the fiber (in the given action)?
I think the answer is no (the first is a subspace of the latter), but I would prefer it to be yes...
If the answer is yes, how do I prove it? could you give me some references?
If the answer is no, could you give me an explicit counterexample?
Thank you very much for the time you dedicated to me and a special thanks to every one who will answer me
best regards
Stgermain
 A: It all works out as well as you could want in every possible sense because of the freeness of the action. As you know, you can make a cover by $G$-stable affine opens, so the real work is in that case.  So we focus on the affine case, and then all hypotheses on the affine can be removed: let $A$ be any ring whatsoever and $G$ a finite group acting freely on $X = $ Spec($A$) in the sense of acting freely on the set of geometric points valued in any algebraically closed field, which forces the strongest sense of acting freely on points valued in any ring at all.   Thus, $G \times X := \coprod_{g \in G} X \rightarrow X \times X$
as functors via $(g,x) \mapsto (x, g.x)$ is a subfunctor inclusion and thus an equivalence relation on $X$ in the sense of functors. Now for the real content: that equivalence relation condition, coupled with $G \times X \rightrightarrows X$ being finite locally free (even finite etale) implies that SGA3, Expose V, 4.1(iv) applies, so $X := {\rm{Spec}}(A)$ is a finite etale cover of $Y := {\rm{Spec}}(A^G)$ and the natural map $$\coprod_ {g \in G} X \rightarrow X \times_ Y X$$ via $x_g \mapsto (x, g.x)$ on the $g$th copy of $X$ is an isomorphism.  This is a deep result of Grothendieck in such generality. In particular, if $A^G$ is a $B$-algebra and $B \rightarrow B'$ is any map of rings then the natural map $$B' \otimes_B A^G \rightarrow (B' \otimes_B A)^G$$ is an isomorphism. 
So the $G$-invariant map $\pi:X \rightarrow Y$ is a $G$-torsor for the etale topology and hence $Y$ is a quotient of $X$ by the $G$-action in all good senses (quotient sheaf, good behavior under base change, universal mapping property, etc.). In particular, by etale descent theory (see the example of "Galois covers" in section 6.2 or so of the book "Neron Models") it follows that $\pi_{\ast}^G$ and $\pi^{\ast}$ induce inverse equivalence between the category of quasi-coherent $O_Y$-modules and $G$-equivariant quasi-coherent $O_X$-modules.  
In particular, for $G$-equivariant quasi-coherent $O_X$-modules $F$, the formation of $\pi_{\ast}(F)$ and $\pi_{\ast}(F)^G$ commute with any base change on the quotient (e.g., passing to a fiber there). Hence, the geometric fibers of $\pi_{\ast}(F)$ are the $G$-invariants on the fibers, as you desired, and more vividly the fiber of $\pi_{\ast}(F)$ at a geometric point $y$ of $Y$ is the direct sum of the fibers at the points in the $G$-orbit fiber of geometric points on $X$ over $y$, with $G$ acting simply transitively on this collection of fibers at points of $X_y$. 
Edit: As t3suji points out, the situation is much simpler (without needing freeness conditions) if $|G|$ is a unit on the schemes in question. 
