Localization of modules; localization and sums commute 
Let $(B_i)_{i\in I}$ be a family of $R$-submodules of a module $B$. Show
$$S^{-1}(\sum_{i\in I}B_i) = \sum_{i\in I} S^{-1}B_i$$
as $S^{-1}R$ submodules of $S^{-1}B$.

If it helps, we know that localization of modules defines an exact functor from $R$-$\mathrm{Mod}$ to $S^{-1}R$-$\mathrm{Mod}$. I've tried to construct module homomorphisms directly but it doesn't seem like that is the best method from what I have been trying. I tried
$$f:S^{-1}(\sum_{i\in I}B_i)\rightarrow \sum_{i\in I} S^{-1}B_i$$
by
$$\frac{(b_i)}{s} \mapsto \left(\frac{b_i}{s}\right)$$
but it doesn't look like the function would be additive. Then trying to construct a function in the opposite direction, it seems difficult to choose some mapping
$$\left(\frac{b_i}{s_i}\right) \mapsto \frac{(\phi(b_i))}{\psi(s_i?)}$$
Can this problem be approached by using localization of modules defines an exact functor?

Exactness of localization can indeed make the problem simpler. The inclusions $B_i \subseteq \sum B_i$ localize to give $S^{-1}B_i \subseteq S^{-1}(\sum B_i)$ for every $i$, so $\sum S^{-1}B_i \subseteq S^{-1}(\sum B_i)$. Conversely, an element of $S^{-1}(\sum B_i)$ is of the form $\frac{b_1 + \cdots + b_n}{s}$, where $b_i \in B_i$. But $\frac{b_1 + \cdots + b_n}{s} = \frac{b_1}{s} + \cdots + \frac{b_n}{s} \in \sum S^{-1}B_i$, so equality holds. 
