Cocomplete abelian category with enough injectives has exact coproducts In this post it is claimed that for any (cocomplete) abelian category with enough injectives, the coproduct functor is exact, that is for a family of short exact sequences $0 \to A_i \to B_i \to C_i \to 0$, also $0 \to \oplus_i A_i \to \oplus_i  B_i \to \oplus_i C_i \to 0$ is exact.
Since taking the coproduct is right exact, it suffices to see that for a family of monomorphisms $A_i \to B_i$ the morphism $\oplus_i A_i \to \oplus_i B_i$ is monic as well.
I thought about this a bit myself but as I’m not very familiar with abelian categories I did not get very far. Any hint, proof or reference for a proof would be very much appreciated.
 A: Let $\mathcal{A}$ be an abelian category with enough injective objects, let $\mathcal{B}$ be the full subcategory of injective objects, and let $[\mathcal{B}, \textbf{Ab}]$ be the category of additive functors $\mathcal{B} \to \textbf{Ab}$.
Then the functor $\mathcal{A}^\textrm{op} \to [\mathcal{B}, \textbf{Ab}]$ defined by $A \mapsto \textrm{Hom} (A, -)$ is exact and fully faithful.
Moreover, it sends coproducts in $\mathcal{A}$ to products in $[\mathcal{B}, \textbf{Ab}]$.
But fully faithful functors are conservative, so this functor also reflects exactness, and $[\mathcal{B}, \textbf{Ab}]$ satisfies axiom AB4* (because $\textbf{Ab}$ does), hence $\mathcal{A}$ satisfies axiom AB4, as claimed.

There is also a somewhat more hands-on argument.
Coproducts are always right exact, so we only have to check that monomorphisms are preserved.
Let $A = \bigoplus_i A_i$, $B = \bigoplus_i B_i$, $C = \bigoplus_i C_i$.
We need to show that the induced $A \to B$ is a monomorphism.
Let $K = \ker (A \to B)$.
It suffices to show that $K \cong 0$.
Since $\mathcal{A}$ has enough injective objects, we may choose an injective object $J$ and a monomorphism $K \to J$.
We then have the following exact sequence:
$$\textrm{Hom} (B, J) \longrightarrow \textrm{Hom} (A, J) \longrightarrow \textrm{Hom} (K, J) \longrightarrow 0$$
On the other hand, for each $i$,
$$0 \longrightarrow \textrm{Hom} (C_i, J) \longrightarrow \textrm{Hom} (B_i, J) \longrightarrow \textrm{Hom} (A_i, J) \longrightarrow 0$$
is exact, so their product is also exact:
$$0 \longrightarrow \prod_i \textrm{Hom} (C_i, J) \longrightarrow \prod_i \textrm{Hom} (B_i, J) \longrightarrow \prod_i \textrm{Hom} (A_i, J) \longrightarrow 0$$
But $\textrm{Hom} (-, J)$ sends coproducts to products, so this implies $\textrm{Hom} (B, J) \to \textrm{Hom} (A, J)$ is an epimorphism, so $\textrm{Hom} (K, J) \cong 0$.
But we have a monomorphism $K \to J$, so this forces $K \cong 0$, as required.
