Given $R$-module maps $\{f_i \colon M_i \to N_i\}_{i \in I}$, the universal properties of the direct sums $\oplus M_i$ and $\oplus N_i$ give a unique map $\oplus f_i \colon \oplus M_i \to \oplus N_i$ such that $M_j \xrightarrow{f_j} N_j \to \oplus N_i$ agrees with $M_j \to \oplus M_j \xrightarrow{\oplus f_j} \oplus N_i$ for each $j \in I$. (The unlabeled maps are the inclusions.)
If we use the explicit constructions of $\oplus M_i$ and $\oplus N_i$ as sets of tuples $\{(m_i)_{i \in I} \mid m_i \in M_i\}$ and $\{(n_i)_{i \in I} \mid n_i \in N_i\}$, then the map $\oplus f_i$ is given explicitly by $(m_i) \mapsto (f(m_i))$. Now it's easy to see that if $f_i$ is an injection for all $i$, then $\oplus f_i$ is, too.
However, I was wondering if there's a way to reason about injectivity of $\oplus f_i$ directly from the universal properties, without using any particular construction for the direct sum.
Best Answer
If $C$ is a category with infinite coproducts, taking $I$-indexed coproducts is a functor $\bigoplus : C^I \to C$ which is left adjoint to the diagonal functor $\Delta : C \to C^I$, and hence which preserves colimits. This implies that it preserves epimorphisms, since those are the morphisms with trivial cokernel pair (in an abelian category substitute "cokernel"). However, there's no abstract nonsense reason it should preserve monomorphisms; those are the morphisms with trivial kernel pair (again, in an abelian category substitute "kernel"), so this would be a limit-commuting-with-colimit type scenario, which usually requires special hypotheses.
Here you can find a counterexample by Zhen Lin in the abelian category $\text{Sh}([0, 1])^{op}$, the opposite of the category of sheaves of abelian groups on the interval.