Short exact sequence of linear representations of a group Let $G$ be a group with $V',V,V''$ being representations of $G$. Let $$0
{\longrightarrow}V'
{\longrightarrow}V\overset{v}{\longrightarrow}V''\longrightarrow
 0\tag{1} $$ be a short exact sequence of $K$-vector spaces and homomorphisms. Then $v$ is a surjection. 
For any representation of G, say V, we define
 $$\pi(x) := \frac{1}{|G|} \sum_{g\in G} g(x)$$   $\forall x\in V$.
and$$ V^G:=\{x\in V:   s.x=x, \forall s\in G\}$$
To show that $\forall x''\in V''^G, \exists x\in V^G$ such that $v(x)=x'' $.
 The proof in the book I am reading is as follows,
$$v(\pi(x))=\pi(v(x)=v(x)=x'')$$
My question is that


*

*Aren't $\pi(x)$ and $\pi(x'')$ in the images of two different $\pi$ functions, one defined on $V$ other on $V''$?

*Why is $\pi$ commuting with $v$?

 A: If the maps in (1) aren't compatible with the repesentations $G \to \mathrm{GL}(V)$, $\mathrm{GL}(V')$, $\mathrm{GL}(V'')$, the result is not true, so in the following, we suppose, that (1) is a short exact sequence of $G$-representations and homomorphisms. 
Serre also writes in the statement of this Corollary (page 61, Corollaire A.3) (emphasis added)

Soit $$ 0 \to V' \to  V \stackrel v\to V'' \to 0 $$
  une suite exacte de représentations de $G$.
  Alors tout élément de $V''^G$ est image d’un élément de $V^G$.

That is, by definition, we have that 
$$ v(g.x) = g.v(x), \quad g \in G, x \in V $$
For any homomorphism of $G$-representations, $\phi\colon V\to W$ note that $\phi$ commutes with the respective $\pi$s. We have for $x \in V$:
\begin{align*}
   (\phi \circ \pi_V)(x) &= \phi\left(\frac 1{|G|}\sum_{g\in G} gx\right)\\
            &= \frac 1{|G|} \sum_{g\in G} \phi(gx)\\
            &= \frac 1{|G|} \sum_{g\in G} g\phi(x) \text{ as $\phi$ is a $G$-homomorphism}\\
            &= (\pi_W \circ \phi)(x)
\end{align*}
Now everything works fine: Let $x'' \in V''^G$, then $\pi_{V''}(x'') = x''$. Choose, as $v$ is onto $x \in V$ with $v(x) = x''$. We have 
$$ v\pi_V(x) = \pi_{V''}\bigl(v(x)\bigr) = \pi_{V''}(x'') = x''. $$
