TITLE: Infinite coproduct of abelian groups
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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.