How to prove $\dim V^{Gl_n}\leqslant \dim(\ker f)^{GL_n} + \dim W^{GL_n}$? Let $V$ and $W$ be vector spaces. Denote by $GL_n:=GL(n, \mathbb K)$ the general linear group of rank $n$ over the field $\mathbb K.$ This group acts regularly on the spaces $V$ and $W.$ Therefore, they have a $GL_n$-module structure.
Assume the homomorphism $f: V\to W$ is an epimorphism of $GL_n$-modules and $V = \ker f \oplus W.$ Then, $\dim V^{Gl_n}\leqslant \dim(\ker f)^{GL_n} +  \dim W^{GL_n},$ where $(V)^{Gl_n}$ denotes the subspace of $V$ consisting of all the invariant classes under the usual action of $GL_n.$ 

How to prove $\dim V^{Gl_n}\leqslant \dim(\ker f)^{GL_n} +  \dim W^{GL_n}$?

Thank for all your help!
 A: The inequality is true for finite dimensional group representations in general, and has nothing to do with any special property of the general linear group.
Let $G$ be any group and $0 \to N \to V \overset{f}{\longrightarrow} W \to 0$ be a short exact sequence of finite dimensional $KG$--modules.  Then
$0 \to N^G \to V^G \overset{f}{\longrightarrow} W^G$, but $V^G \overset{f}{\longrightarrow} W^G$ is not necessarily surjective.  So $\dim V^G \le \dim N^G + \dim W^G$, with equality precisely when $V^G \overset{f}{\longrightarrow} W^G$ is surjective. 
Edit:  Example of failure of surjectivity: Let's make an example with the field $K = \mathbb R$ and $G = (\mathbb R, +)$.  Take $V = \mathbb R^2$, and let $G$ act by $\phi(t) = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$. Let $N = \mathbb R e_1$. Check $N = V^G$.   Let $W = V/N = \mathbb R (e_2 + N)$.   Check $W = W^G$.   We have
$0 \to N \to V \overset{f}{\longrightarrow} W \to 0$, where $f: V \to V/N = W$ is the quotient map.  Note that
 $f(V^G) = (0) \ne W^G$.   
Turn this into an example with $G = \operatorname{GL}_1(\mathbb R)$ by taking
$ s \mapsto  \phi(\log(s^2))$.
There is a fancy terminology related to this.  $V \mapsto V^G$ is a functor.  This functor is left exact but not exact.
