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 ?


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.

  • $\begingroup$ I'm surprised, how can an equivalence of categories not preserve a coproduct ? (I'm answering this in my head as I'm writing the comment : the coproduct in the subcategory will not be a coproduct in $R$-Mod...) Thank you very much !! $\endgroup$ Jun 24 '18 at 17:46
  • $\begingroup$ I actually needed this lemma to prove that homology commutes with direct sum in a general abelian category; but couldn't prove it with Freyd's theorem (and my proof with Freyd's theorem was wrong as you pointed out) - and it turns out that all of this is actually very related and I found a MO-talk where it was discussed and it turns out that homology does not, in general, commute with direct sums $\endgroup$ Jun 24 '18 at 18:07
  • $\begingroup$ As a general rule of thumb, any statement about infinite (co)limits in an abelian category which does not easily follow formally from the definitions is probably false in general. $\endgroup$ Jun 24 '18 at 18:11

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