Let $R$ and $R'$ be rings. Let $\mathcal{F}:R - \text{Mod} \to R' - \text{Mod}$ be an equivalence of categories with quasi-inverse $\mathcal{G}$. Let $V$ be an indecomposable $R$-module.
Claim: $\mathcal{F}(V)$ is an indecomposable $R'$-module.
Proof: Let $\mathcal{F}(V) = U' \oplus W'$ for two $R'$-modules $U'$ and $W'$. We want to show either $U' = 0$ or $W' = 0$. Since $\mathcal{F}$ is an equivalence of categories, the image is dense in $R' - \text{Mod}$ so there exists $R$-modules $U$ and $W$ with $\mathcal{F}(U) \cong U'$ and $\mathcal{F}(W) \cong W'$. Then, since $\mathcal{F}$ is an equivalence of categories and thus additive, we have
$$\mathcal{F}(V) \cong \mathcal{F}(U) \oplus \mathcal{F}(W) \cong \mathcal{F}(U \oplus W).$$
Applying the quasi-inverse $\mathcal{G}$ we find
$$V \cong \mathcal{G}(\mathcal{F}(V)) \cong \mathcal{G}(\mathcal{F}(U \oplus W)) \cong U \oplus W.$$
Since $V$ is an indecomposable $R$-module, either $U$ or $W$ is the zero module. Thus either $U' \cong \mathcal{F}(U)$ or $W' \cong \mathcal{F}(W)$ is the zero module. Here we are using the fact that $\mathcal{F}$ is additive again and thus takes a $0$ element in $R - \text{Mod}$ to a $0$ element in $R' - \text{Mod}$.
My Question: Is the above proof valid? Am I missing a hidden justification any where? Is there a faster/slicker/more enlightening way to prove the above statement.