exists homomorphism of $G$-representations $\pi: V \to V$ with image $X$ 
Let $G$ be a group (not necessarily finite) and $F$ a field. Let $V$ be a $G$-representation. Suppose $V$ is isomorphic to a direct sum of irreducible representations. Let $X \subset V$ be any subrepresentation. Prove that there exists a homomorphism of $G$-representations $\pi: V \to V$ with image $X$.

I'm kinda confused on how to start this problem. I have that any subrepresentation of $V \oplus V$ is of the form $\{0\} \oplus V$ or $V \oplus \{0\}$, but I'm not sure what to do next. Any help would be greatly appreciated.
 A: The statement that any subrepresentation of $V \oplus V$ is of the form $\{0\} \oplus V$ or $V \oplus \{0\}$ is false; a counterexample is given by $\{(v, v)\text{ }|\text{ }v \in V\}$.
Write $V = \bigoplus_i U_i$ with $U_i$ irreducible. We claim there exists $Y \subset V$ with $V = X \oplus Y$.
Proceed by contradiction, and choose $X \subset V$ of maximal dimension so that the claimed result is false. Clearly $X \neq V$ $($otherwise we could take $Y = 0)$. There exists $i$ so that $U_i$ is not contained in $X_i$; by irreducibility of $U_i$, the intersection $U_i \cap X = \{0\}$. Write $X' = X + U_i$; it is the direct sum of $X$ and $U_i$:$$X' = X \oplus U_i.$$By assumption $(X$ is a counterexample of maximal dimension$)$ there is $Y'$ so that$$V = X' \oplus Y'.$$Then $V$ is the direct sum of $X$, $U_i$, and $Y'$; in particular,$$V = X \oplus (U_i \oplus Y'),$$contradiction.
Thus there exists $Y \subset V$ with $V = X \oplus Y$. The map $\pi: V \to V$ defined by $x + y \mapsto x$ $(x \in X$, $y \in Y)$ is a $G$-homomorphism with image equal to $X$, as desired.
