Proof that $~\hom_G(V,\bigoplus U)=\bigoplus\hom_G(V,U)$ Yesterday I have already asked for help in this post to prove that $~\hom_G(V,\bigoplus U)=\bigoplus\hom_G(V,U)$ for finite $\bigoplus$. Unfortunately, I did not fully understand the answer because it involved category theory and a universal property.
Would the following proof also be correct?
Thank you very much for your help!

We are required to prove that
\begin{equation}\text{Hom}_G(V, W_1\oplus W_2)= \text{Hom}_G(V,W_1) \oplus~\text{Hom}_G(V,W_2),\end{equation}
where $V$ is irreducible.
Let $T \in \text{Hom}_G(V,W_1) \oplus~\text{Hom}_G(V,W_2)$. Hence $T = T_1 + T_2$, where the combination $T_1 \in \text{Hom}_G(V,W_1)$ and $T_2 \in \text{Hom}_G(V,W_2)$ is unique.
Let $v \in V$. Thus $Tv = (T_1 + T_2)v = T_1v + T_2v = w_1 + w_2$, where $w_1 \in W_1$ and $w_2 \in W_2$. Hence $T \in \text{Hom}_G(V, W_1 + W_2)$.
So we still have to prove that $T \in \text{Hom}_G(V, W_1\oplus W_2)$. To this end, let us suppose $w \in W_1 \cap W_2$.
Let $(v_1, v_2, ..., v_n)$ be a basis for $V$. Let us define $Sv_1 = w, Sv_i = 0$ for all $i \neq 1$. Hence $S \in \text{Hom}_G(V,W_1) \cap~\text{Hom}_G(V,W_2)$. But then $S = 0$, so $Sv_1 = w = 0$. Hence $W_1 \cap W_2 = \left\{0\right\}$.
Thus, $T \in \text{Hom}_G(V, W_1\oplus W_2)$ and
\begin{equation}\text{Hom}_G(V, W_1\oplus W_2) \supset \text{Hom}_G(V,W_1) \oplus~\text{Hom}_G(V,W_2).\end{equation}
To prove the other direction, let now $T \in \text{Hom}_G(V, W_1\oplus W_2)$.
Hence $Tv = w_1 + w_2$ for all $v \in V$, where $w_1 \in W_1, w_2 \in W_2$ are unique for each $v$.
Let $T = A + B$, where we let $Av = P_{W_1,W_2}Tv$ and $Bv = P_{W_2,W_1}Tv$, where $P_{W_i,W_j}$ denotes the projection onto $W_i$ with null space $W_j$. Thus, $A \in \text{Hom}_G(V,W_1)$ and $B \in \text{Hom}_G(V,W_2)$, so $T \in \text{Hom}_G(V,W_1) + \text{Hom}_G(V,W_2)$.
For $T \in \text{Hom}_G(V,W_1) \cap \text{Hom}_G(V,W_2)$, $Tv \in W_1$, $Tv \in W_2$ for all $v$, so $Tv \in \left\{0\right\}$, hence $T = 0$.
This proves that $T \in \text{Hom}_G(V,W_1) \oplus~\text{Hom}_G(V,W_2)$, so
\begin{equation}\text{Hom}_G(V, W_1\oplus W_2) \subset \text{Hom}_G(V,W_1) \oplus~\text{Hom}_G(V,W_2),\end{equation}
which proves the result, as required.
Thus, by mathematical induction, $~\hom_G(V,\bigoplus U)=\bigoplus\hom_G(V,U)$, as required.
 A: Too long; didn't read. Nothing needs to be irreducible, the direct sum doesn't even need to be finite.
We can work in the generality of $A$-modules (since a linear representation of a group is just a module over the group algebra). Suppose $A$ is a $k$-algebra, where $k$ is a field, so the homs are vector spaces over $k$ and the isomorphism between them is as vector spaces.
The isomorphism of homs should ultimately be "obvious." Any element of $U$ looks like a finite sum of the form $\sum_i u_i$ with $u_i\in U_i$ for each $i$. Thus if $\phi:V\to U$, for every $v\in V$ since $\phi(v)\in U$ we have a unique decomposition $\phi(v)=\sum_i\phi(v)_i$ where $\phi(v)_i\in U_i$ for each $i$. Thus, $\phi:V\to U$ can be decomposed into these coordinate maps $v\mapsto\phi(v)_i$. One easily checks that
$$(\lambda\cdot\phi)(v)_i=\lambda\cdot\phi(v)_i \quad \phi(av)_i=a\phi(v)_i \quad (\phi+\psi)(v)_i=\phi(v)_i+\psi(v)_i$$
for all $v\in V$, $\lambda\in k$, $a\in A$ and any two $\phi,\psi\in\hom_A(V,U)$. Hence the map $\phi\mapsto(\phi_i)$ has domain and codomain $\hom_A(V,U)\to\bigoplus_i\hom_A(V,U_i)$. The inverse map in the other direction is simply $(\phi_i)\mapsto\sum_i\phi_i$. Thus we have a pair of mutually inverse linear maps between the homs.
