For two invariant complements in the space of representation $T$. Prove that $T_{W_1}$ isomorphic to$ T_{W_2}$ Let $W_{1}$ and $W_{2}$ be two invariant complements of the invariant space $U$ in the space of the representation $T$ prove that $T_{W_{1}}$ equivalent to $T_{W_{2}}$.
Invariant Complement Definition:

Two representations are equivalent iff:

Could anyone give me a hint for doing so please?
Thank you 
 A: Let $V=U\oplus W$. Define the quotient module $V/U$ by $g\cdot (v+U)=g\cdot v+U$ for $v\in V$ and $g\in G$. Can you show that this indeed defines a module? Then use the canonical quotient map $q:V=U\oplus W\rightarrow V/U$ and restrict it to $W$. Then $q\mid_W$ is injective since $W\cap U=\{0\}$ and surjective by the fact that $V=U+W$. Also observe that $q(g\cdot w)=g\cdot w+U =g\cdot(w+U)=g\cdot q(w)$ so $q$ is a $G$-module morphism. Hence $q$ is an isomorphism between $G$-modules. As a result any $G$-invariant complement of $U$ in $V$ is isomorphic to the module $V/U$ which implies that they are all in fact isomorphic as $G$-modules.
Edit : To clarify, I would like to explain a bit more. For a submodule $U$ of $V$, you can define the quotient module $V/U$ as a I explained above. Now, if $W$ is also a submodule of $V$ satisfying $V=U\oplus W$, the quotient map $q:V\rightarrow V/U$ will be injective when restricted to $W$ (if $q(w)=0+U$, then $w+U=0+U$ so $w\in U$, and since $U\cap W=\{0\}$ we get $w=0$. Hence $q\mid_W$ is injective). Moreover, it will be surjective since for every element $v+U$ in $V/U$ we can write $v=w+u$ for some $w\in W$ and $u\in U$ (this is true because $V=U+W$) and then $v+U=w+u+U=w+U=q(w)$. Lastly, it is a $G$-module morphism as I have shown above. Then we deduce that $q\mid_W$ is in fact an isomorphism of two $G$-modules, $W$ and $V/U$. Now this is true for every submodule $W$ of $V$ satisfying $V=U\oplus W$. Hence if both $W_1$ and $W_2$ satisfy this property, we conclude that they are both isomorphic to $V/U$ and thus, they are in fact isomorphic to each other.
