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I thought about the problem of how to understand coproduct and direct sum and I think this could be thought of as a way of thinking. I am posting this to verify if my understanding is correct.

So what we need is the following.

Given data: $$\begin{align} &f_i\colon X_i\to P \\ &g_i\colon X_i \to X \end{align}$$ Now define in someway $$g\colon P \to X$$

$P$ is the coproduct we are trying to define. Please note that we need to define $g$ given $g_i$. $g$ is actually to be defined so that it completes the commutative diagram and then only $P$ is the coproduct.

The most obvious way of defining $g$ on each of the basis elements is $g := \sum g_i$ where $i$ ranges over the index set.

Now all our group, ring homomorphism operations are defined over finite sums or finite products. So there is no way we could make sense of an infinite sum in the definition of $g$. Thus we have to restrict it to finite sum. Hence if we have to restrict it to finite sum then only a finite number of elements of $P$ could be non zero and the rest are all $0$s.

This is my understanding and if somebody could point out that there is some gaps it would be very helpful.

Thanks

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    $\sum g_i$ does not make sense in arbitrary categories; are you assuming that you are in an abelian category? Also, how can you be "given" the $fi$, and yet be "trying to define $P$"? You cannot be "given" a function into an object that has not been defined.2011-08-15
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    Yes I am assuming abelian category. Furthermore, I understand your statement about P but I am really not sure how to word it. I want to somehow understand why is it so that if u have a coproduct you need to have a finite number of non zero elements and cannot have an infinite number of non zero elements in P. Again I am not sure if I am able to explain properly. But (1,1, 1, ... ) is not a member of P but (1, 1, 0, ....) is a member. So this is the major motivation. So if you could tell me how the post should be modified it would be helpful.2011-08-15
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    I am not sure but should it be worded as how do u define g then. Because the direct sum could be thought of as the ( P, g_{i}, g) all these together. Loosely P is called the direct sum but actually it should be the three together if I am not highly mistaken. The probably if that is true then I would edit the post suitably again.2011-08-15
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    No, $g$ is induced by the universal property of the coproduct, and is not *part* of the coproduct. Also, you aren't just considering an abelian category, you are considering a category in which the underlying set of the product is the product of the underlying sets and the underlying set of the coproduct is the restricted direct product. I don't know if this always holds. But since it is clear you are confused, so am I, so I cannot tell you how to "unconfuse" your question.2011-08-16
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    "finite number of non zero elements": http://math.stackexchange.com/questions/523670/infinite-coproduct-of-abelian-groups2014-05-01

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