Projections from a coproduct in an abelian category Suppose you have an abelian (perhaps less is needed, additive, preadditive ? ) category $C$ with arbitrary (small) coproducts.
Given a small family $(C_i)_{i\in I}$ of objects one may define "projection maps" $\pi_j:\displaystyle\bigoplus_{i\in I}C_i \to C_j$ simply by applying the universal property to $id_{C_j}$ and $0:C_i\to C_j$ for $i\neq j$.
In familiar abelian categories (categories of modules)  these $\pi_j$ "completely determine" everything : for instance if $\pi_j\circ f = \pi_j\circ g$ for all $j$, then $f=g$.
As a consequence, because of Freyd's embedding theorem, this result holds in $C$: if $\forall j, \pi_j\circ f = 0$, then $f=0$ (as usual, pick the smallest abelian category containing all the relevant stuff, it will be small, apply Freyd's theorem, conclude)
I was wondering if there was a more intrinsic proof of this - there probably is, so I was rather wondering what such a proof may be.

What would a category theoretic proof of this result look like ? 

 A: This is not true in general!  You can't use the embedding theorem to prove it, because the embedding you obtain may not preserve infinite coproducts.
Indeed, the category $Ab^{op}$ is a counterexample.  In terms of abelian groups, your statement for $Ab^{op}$ would be saying that if $(C_i)$ is a family of abelian groups then a map out of $\prod C_i$ is determined by its composition with all the inclusions $C_j\to \prod C_i$.  But this isn't true, because the images of these inclusions only generate the direct sum $\bigoplus C_i$, not the whole direct product.  So for instance, the quotient map $\prod C_i\to \prod C_i/\bigoplus C_i$ is $0$ when composed with each inclusion but is not the zero map (as long as infinitely many of the $C_i$ are nontrivial).
More generally, your condition is equivalent to saying that the canonical morphism $\bigoplus_{i\in I} C_i\to \prod_{i\in I}C_i$ is a monomorphism, assuming the product exists.  For categories of modules (and most other familiar examples), this morphism is a monomorphism, but not an epimorphism, meaning that your condition will fail in the opposite category.
