Can we derive this property of the direct sum from the categorical definition of coproduct?

Question

I know that, for an infinite family $\{B_i\}$ of abelian groups, the direct sum is a subgroup of the direct product$-$ $$ \bigoplus_i B_i \subseteq \prod_i B_i $$ where an element $(b_i)$ of the direct sum looks just like an element of the direct product, with the condition that cofinitely many $b_i$ are zero. Now, from the group-theoretic perspective this makes perfect sense to me. However, I'm wondering if there's a way to explain why the coproduct exhibits this cofiniteness property, while the product does not, just by looking at the relevant diagrams:

          

Thanks for any help in understanding this!

Edit: I have a vague intuition this is related to there being many morphisms going into the direct sum, which in some sense "obfuscates" the information and requires us to fall back on what we know about sums from the definition, which imposes the cofiniteness condition. Am I on the right track here?

Answer 1

I think the key here is that the category of abelian groups (like the category of $R$-modules, and other categories) has a zero object.

Definition. Let $\mathbf{C}$ be a category. A zero object in $\mathbf{C}$ is an object $\mathbf{0}$ that is both initial and terminal in $\mathbf{C}$; that is, such that for every object $C$ of $\mathbf{C}$, there is a unique morphism $\mathbf{0}_{\mathbf{0}C}\colon \mathbf{0}\to C$ and a unique morphism $\mathbf{0}_{C\mathbf{0}}\colon C\to\mathbf{0}$.

The existence of a zero object leads to the existence of “canonical morphisms” between any two objects:

Proposition. Let $\mathbf{C}$ be a category with a zero object $\mathbf{0}$. Then for any $X,Y\in\mathrm{Ob}(\mathbf{C})$, we have a “zero morphism” $\mathbf{0}_{XY}\colon X\to Y$, defined by $\mathbf{0}_{XY} = \mathbf{0}_{\mathbf{0}Y}\circ\mathbf{0}_{X\mathbf{0}}$. These maps have the property that for every $X,Y,Z\in\mathrm{Ob}(\mathbf{C})$, $\mathbf{0}_{XZ} = \mathbf{0}_{YZ}\circ\mathbf{0}_{XY}$.

For example, $\mathsf{Ab}$ has a zero object (as does $\mathsf{Group}$): the trivial group.

Say you have a category with a zero object. If $\{X_i\}$ is a family that has both a product $P$ (with projections $p_i$) and a coproduct $Q$ (with coprojections $q_i$), then you get a “canonical” morphism $Q\to P$: given $i\in I$, we can map $f_i\colon X_i\to P$ by the map induced by family $f_{ij}\colon X_i\to X_j$, with $f_{ij}=\mathrm{id}_{X_i}$ if $i=j$, and $f_{ij}=\mathbf{0}_{X_iX_j}$ if $i\neq j$; the map $f_i$ satisfies that $p_j\circ f_i$ is the zero map if $j\neq i$, and the identity map if $i=j$.

Then the universal property of $Q$ means that the family of maps $f_i\colon X_i\to P$ induce a map $f\colon Q\to P$ with the property that $f_i=f\circ q_i$ for each $i$. In particular, you get that $p_j\circ f\circ q_i$ is the zero map if $i\neq j$, and is the identity if $i=j$.

In the case of finitary algebras (in the sense of Universal Algebra), where the product has underlying set the cartesian product of the underlying sets, if there is a zero object (there isn’t always; $\mathsf{R}^1$, rings with unity, does not have a zero object; neither does $\mathsf{Semigroup}$), the subobject $f(Q)$ of $P$ has the property that for each $\mathbf{x}\in f(Q)$, $p_i(\mathbf{x})$ lies in the image of the zero object in $X_i$ for all but finitely many $i$ (this can be proven by induction on the length of the terms in the category). But we are already in a rather restricted class of categories: categories of algebras with zero objects.

Note that this “canonical map” $Q\to P$ in categories with zero objects need not be an embedding. For example, in the category of all groups, $\mathsf{Group}$, $Q$ is the free product of the $X_i$ and $P$ is the cartesian product. The map $Q\to P$ has nontrivial kernel, generated as a normal subgroup by the subgroups $[q_i(X_i),q_j(X_j)]$ with $i\neq j$; this nontrivial normal subgroup is called the “cartesian of $Q$”. But in the category of groups, it is still true that any element in the image of this morphism has only finitely many nontrivial coordinates, because the elements of $Q$ are “finitary”: they involve only finitely many elements from finitely many of the $X_i$.

Answer 2

There is no purely categorical argument for the existence of a canonical morphism from a coproduct to a product since then we could dualize the argument and get a canonical morphism from a produt to a coproduct (and we know that such a morphism doesn't exist e.g. in the category of Abelian groups). You have to use the Abelian group structure in some way: in this case, you have to use the fact that canonical projections $\bigoplus_i B_i \to B_i$ exist, so the universal property of $\prod_i B_i$ yields a canonical map $\bigoplus_i B_i \to \prod_i B_i$.