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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:

product diagram           coproduct diagram

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?

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  • $\begingroup$ I’m unclear what you mean by “cofiniteness condition”. $\endgroup$ Apr 2 at 20:19
  • $\begingroup$ The direct sum is not the categorical coproduct. The coproduct for groups is the "free product", or "amalgamated sum". $\endgroup$
    – Couchy
    Apr 2 at 20:20
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    $\begingroup$ @couchy For Abelian groups it is. $\endgroup$ Apr 2 at 20:30
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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 $X_i$.

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  • $\begingroup$ Thanks Arturo for the thorough explanation! Could you give a bit more detail on the induction step? What do you mean "length of the terms in the category"? $\endgroup$
    – albert
    Apr 5 at 15:59
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    $\begingroup$ @albert: The notion of “term” depends on the signature of the algebras in question. Essentially, they are all expressions you can build up from constant terms and from the operations in the algebra. For groups, you would have all expressions involving products, inverses, and the identity element, of variables, without doing any reductions whatsoever. The length is then defined in terms of the constants and operations. But the important point is that because every operation has only finite arity, all of these expressions are finite, so they will only involve finitely many things. $\endgroup$ Apr 5 at 17:14
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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 $\bigoplus_i B_i \to \prod_i B_i$.

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  • $\begingroup$ Sorry, my question was a bit vague. I'm asking about cofiniteness of zeroes in the direct sum, I added some more details $\endgroup$
    – albert
    Apr 2 at 20:13
  • $\begingroup$ Or you can use the fact that we have a zero object, so that you get embeddings $B_j\to\prod B_i$ to get the canonical map. $\endgroup$ Apr 2 at 20:15
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    $\begingroup$ This is incorrect. There is in fact always a canonical morphism from the coproduct to the product (induced by the identity maps via the universal properties of products and coproducts) in any category with a zero object. There is no way to "dualize" this to get a map from the product to the coproduct. $\endgroup$ Apr 2 at 20:15
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    $\begingroup$ @diracdeltafunk: I think you need a zero object for that. What is the canonical morphism from the disjoint union of two sets into their cartesian product, in $\mathsf{Set}$? $\endgroup$ Apr 2 at 20:17
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    $\begingroup$ Thanks! Realized my mistake right after I commented. But I would still call this "purely categorical" -- it has nothing to do with the fact that these objects are abelian groups. $\endgroup$ Apr 2 at 20:19

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