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One can see on every text (book, lesson, comments) that a direct sum/coproduct of abelian groups is the same as a finite product but in the infinite case, the direct sum/coproduct is only a subgroup of the product.

Why does an element in an infinite direct sum only have finitely many components different from the neutral elements?

I don't see where the "finite" comes from when I look at the universal property.

Also, in another situation, a condition of "finiteness" appears and I think that is because of the same structure: a freely generated group, vectors space, algebra etc... has elements that are finite product/sum of generating elements. Because, if one really wants an infinite "sum" (or sequence) one may just take a formal serie. That is a ring, hence also an abelian group.

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    $\begingroup$ It exists. It is the infinite direct sum, of course. $\endgroup$ – Zhen Lin Oct 12 '13 at 16:45
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As far as I can tell, the other answers didn't respond to

I don't see where the "finite" comes from when I look at the universal property.

So the question is: Given that $P$ is a coproduct of the abelian groups $A_i$, how can we deduce - from the universal property, not from the construction of $P$ - that every element of $P$ is a finite sum of elements coming from the $A_i$?

Well, first of all, it doesn't make sense to consider infinite sums, because abelian groups don't have such an operation.

Now, for the proof, let $Q$ be the subgroup of $P$ generated by the images of the $A_i$. Then the universal property of $P$ induces a canonical homomorphism $P \to Q$, and one checks that it is inverse to the inclusion $Q \hookrightarrow P$. It follows that $P = Q$. Hence, every element of $P$ lies in $Q$, hence is a finite sum of elements coming from the $A_i$. Note that we didn't use the construction of $P$, only its universal property.

In other categories coproducts can contain "infinite" sums. Take for instance the category of Banach spaces, with continuous linear maps.

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A coproduct of a family $A_i$ of abelian groups is, by definition, an abelian group $B$ together with a family of morphisms $f_i\colon A_i\to B$ such that, for any family of morphisms $g_i\colon A_i\to C$, there exists a unique $\varphi\colon B\to C$ with $\varphi\circ f_i=g_i$.

So we need a way to map each $A_i$ into some group $B$ and we might consider the product $\prod_i A_i$, with the obvious maps. However, when we want to build the morphism $\varphi$, we see that we are at stake in defining its action on an arbitrary family $(a_i)\in\prod_i A_i$ because we don't know how to sum an arbitrary family. But we do know how to sum a finite number of terms: so, why not take only the elements in $\prod_i A_i$ having only a finite number of non zero terms? Hey! This seems a pretty good idea! Define the set as $\bigoplus_i A_i$ and $\varphi\colon\bigoplus_i A_i\to C$ by $$ \varphi(a_i)=\sum_i g_i(a_i) $$ which is well defined because we can throw away the zero terms.

More precisely, the set $\bigoplus_i A_i$ consists of the families $a=(a_i)_{i\in I}$ such that $a_i=0$ for all but a finite number of indices. In other words, an element $(a_i)_{i\in I}\in\prod_iA_i$ is declared to belong to $\bigoplus_iA_i$ if and only if $\{i\in I: a_i\ne0\}$ is finite.

Now it's trivial to verify that this is indeed a morphism and verifying that $\bigoplus_i A_i$ is a subgroup of $\prod_i A_i$ is also easy.

So we have our coproduct.

If you want to think in terms of series, consider those that converge no matter what topology you endow the reals with: they are just the series with finitely many non zero terms.

Note that this intuition fails in other situations: the coproduct of a family of groups is very different from the product, even with only two factors. See Coproducts exist in the category of groups and note that the free product of two non trivial groups is infinite.

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  • $\begingroup$ You didn't proof that it satisfies the universal property: because you are secretely taking a finite family $g_i$ and then define $\phi \equiv \sum_i g_i$, whereas the universal property says: "for any family"... $\endgroup$ – Noix07 Oct 14 '13 at 12:55
  • $\begingroup$ @user39158: No, egreg proves the universal property for arbitrary families. $\endgroup$ – Martin Brandenburg Oct 14 '13 at 13:14
  • $\begingroup$ aren't you struck by the fact that if we have an infinite family $g_i$ but only finitely many $f_i$ are non zero when applied to a certain sequence $a_i$, then we should have $\phi\circ f_j(a_j)= \phi (0) = g_j(a_j)$, for j such that $f_j(a_j) =0$? $\endgroup$ – Noix07 Oct 14 '13 at 13:47
  • $\begingroup$ @user39158 Some work on details has to be done for defining the sum over any family (with at most finitely many non zero elements), but that's just details. $\endgroup$ – egreg Oct 14 '13 at 13:47
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    $\begingroup$ @egret, you are right. We can have $\phi\circ f_i=g_i$ for all $i$ even when there are infinitely many non-zero $f_i$, $g_i$ and the domain of $\phi$ is $\bigoplus A_i$ only. $\endgroup$ – velut luna Jan 29 '15 at 17:32
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(complete edit of an old answer...following the notation of egreg.)

If the coproduct $B$ exists together with the canonical maps $f_i : A_i \rightarrow B$ then given an arbitrary family of functions $g_i : A_i \rightarrow C$ the map $\phi$ is defined by the universal property only on $\bigcup_i f(A_i) \subset B$ by $$\phi \circ f_i = g_i \ ,\quad \forall\ i $$

$\bigcup_i f(A_i)$ is not a group but the universal property is actually defining the "smallest" group that contains it, namely $B$. I don't know if it is true but as a guess I would say that it is also the freely generated abelian group on the $A_i$ divided by the relations on the $A_i$ and this is why it is the elements in the product that have only finitely many non trivial components

On the other hand the product (category of abelian groups) doesn't satisfy the universal property of the coproduct for an arbitrary family because a map $$ \tilde{\phi}: \Pi_i A_i \longrightarrow C $$ is not uniquely defined by the conditions $\ \tilde{\phi} \circ f_i = g_i \ ,\quad \forall\ i$

Indeed there is a choice of value for any $(a_i)\in \Pi_i A_i$ with infinitely many non trivial components.

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The example from calculus requires some notion of convergence to make sense of the infinite sum. That isn't available in abelian groups in general. The product of a sequence of abelian groups $A_i$, $i \in \mathbb{N}$ comprises all sequences $a_i$ such that $a_i \in A_i$ for all $i$, with addition defined pointwise: $(a+b)_i = a_i + b_i$. The coproduct or direct sum is the subset of the product comprising the sequences $a_i$ where $a_i = 0$ for all but finitely many $i$, so that the convergence issue does not apply.

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  • $\begingroup$ I don't want people to repeat what is already writen every time one talks about direct sum. There is no "finite" in the definition of coproduct $\endgroup$ – Noix07 Oct 12 '13 at 16:58
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    $\begingroup$ The coproduct of an infinite family of abelian groups is the subgroup of the product comprising the finitely non-zero elements. "Finite" is not part of the the defining property of a coproduct, but is required to describe the construction of the coproduct. (What I wrote is not covered by the first paragraph of your question, so I find the first sentence of your comment ungrateful, rude and irrelevant.) $\endgroup$ – Rob Arthan Oct 12 '13 at 19:59
  • $\begingroup$ Ok it was rude. But as a student it is not possible to ask a simple question and have the relevant answer when one doesn't show that he is a good student, otherwise, the teacher will always get back to the basis definition that doesn't answer the question. It is the most striking feature that in our case product and coproduct coincide for finite famlies and don't for infinite, and if I was to write a lesson on this, the first remark I would write is why it is so $\endgroup$ – Noix07 Oct 14 '13 at 13:38

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