Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


share|improve this question
$\sum g_i$ 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. –  Arturo Magidin Aug 15 '11 at 20:05
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. –  sumo Aug 15 '11 at 20:58
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. –  sumo Aug 15 '11 at 21:03
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. –  Arturo Magidin Aug 16 '11 at 3:13
"finite number of non zero elements": math.stackexchange.com/questions/523670/… –  user39158 May 1 at 14:05

2 Answers 2

I saw you posted on MO, and it seems you are trying to relate the ideas of coproduct and direct sum. Put simply, the notion of a coproduct is a generalization of the notion of a direct sum to a general category. For concreteness, consider the case of vector spaces over a field $k$. Given a $k$-vector space $W$ and two subspaces $U, V$, the sum $U+V$ is just $\{u+v|u\in{}U,v\in{}V\}$, and the sum is said to be direct if we have unique expression, i.e. if $u,u'\in{}U$, $v,v'\in{}V$ and $u+v=u'+v'$, then $u=u'$ and $v=v'$.

Now, given a direct sum $U\oplus{}V$, there are 'injections' $i:U\to{}U\oplus{}V$ and $j:V\to{}U\oplus{}V$ taking elements to themselves, and we have the following universal mapping property:

Given a $k$-vector space $T$ and linear maps $f:U\to{}T$, $g:V\to{}T$, there is a unique linear map $[f,g]:U\oplus{}V\to{}T$ such that $[f,g]i=f$ and $[f,g]j=g$.

The map $[f,g]$ is defined by taking $u+v$ to $f(u)+g(v)$. This is well-defined, since the sum is direct. By the usual arguments, we can show that this universal property characterizes $U\oplus{}V$ up to isomorphism, so we have an abstract characterization of a direct sum purely in terms of maps. Of course this characterization just says that $U\oplus{}V$, together with $i$ and $j$, is a coproduct diagram of $U$ and $V$ in $k$-Vect.

These remarks generalize to infinite direct sums as well, that is, given a family $(U_i)_{i\in{}I}$ of subspaces of $W$, the direct sum of these spaces, together with the relevant injection maps, is a coproduct diagram in $k$-Vect.

In a general category, it doesn't usually make sense to talk about elements and maps as we have above, but it always does make sense to talk about morphisms, diagrams, and universal properties. So, the notion of a coproduct really generalizes direct sums to the general categorical setting.

I hope this was helpful.

share|improve this answer

Perhaps this might work: you understand that in an abelian category the coproduct and the product of two objects are the same (when they both exist). You want to "understand" something about the coproduct and the product of infinite families and how they may differ.

In an abelian category, if $J\subseteq K$ are finite, then you have maps from $\prod_{j\in J}X_j$ to $\prod_{k\in K}X_k$ and back; the map from the $J$-indexed product to the $K$-indexed product is obtained by "extending by $0$s", while the map back is obtained by dropping the components not in $J$. These are in fact maps induced by the universal properties of the product and the coproduct by the canonical projections and immersions, recalling that the finite products are equal to the finite coproducts.

That means that given an infinite family $\{X_i\}_{i\in I}$, you can look at the family of finite products $\bigl\{\prod_{j\in J}X_j\mid J\subseteq I,\ |J|\lt\infty\bigr\}$ as both a directed family (ordered by the partial order of inclusion among finite subsets) and as an inversely directed family. If you take the direct limit of this family and interpret the objects as coproducts, then you get the coproduct of the full infinite family. If you take the inverse limit of this family and interpret the objects as products, you get the product of the full infinite family.

The direct limit of the objects is defined by "local" conditions, so that intuitively you only need to worry about what is happening "locally", hence the objects should only require finitely many information; whereas the inverse limit requires a whole family of consistent objects, so that you need "a lot more" information. Intuitively, local information just isn't enough.

For abelian groups, modules, and vector spaces, the directed limit consists of equivalence classes of objects of the form $[(x,J)]$ with $x\in \prod_{j\in J}X_j$, with $[(x,J)]=[(y,K)]$ if and only if there exists $M$ finite, $J,K\subseteq M$, and such that the images of $x$ and $y$ in $\prod_{m\in M}X_m$ agree; so elements are completely determined by information on a finite subset of $I$.

On the other hand, the inverse limit is constructed at the set level as a subset of the set-theoretic product $\times(\prod X_i)$ of the families, taking those elements $(x_J)$ indexed by the finite subsets of $I$, such that if $J\subseteq K$ then the image of $x_K$ in $\prod_{j\in J}X_j$ is precisely $x_J$ ("consistent families"). This means that you may need information about what is happening on all coordinates to determine an element, and there are elements that are not obtained merely by local information.

(A possible model might be the difference between the directed limit of the cyclic groups $\mathbb{Z}_{p^n}$ and the inverse limit of the same groups; the directed limit is the Prüfer group, in which each element is, in a sense, "finite", whereas the inverse limit is the $p$-adic integers, in which you have elements that are not "finite", but rather "come" from 'infinite tuples'.)

share|improve this answer
Thx so much for clarifying. It was very very helpful. This was the thing which clarified everything. –  sumo Aug 16 '11 at 5:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.