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 have the definition of a coproduct which is as follows:

A coproduct of $\{A_\alpha\}$ in $\mathcal{G}$, where $\mathcal{G}$ is a category and $\{A_\alpha\}$ a collection of objects, is an object $Q$ of $\mathcal{G}$ with $\pi_\alpha:A_\alpha\rightarrow Q$ s.t. $\forall{C}\in obj\{\mathcal{G}\}$ with morphisms $\phi_\alpha:A_\alpha\rightarrow C$ s.t $\exists! f:Q\rightarrow C$ s.t. $\phi_\alpha=f\circ\pi_\alpha$

Now I have been given the example with if our objects are abelian groups and morphisms are homomorphisms then this defines the direct sum. I was wondering how we go about proving this? I can kind of see why it does but how do we show for example that nothing else but the direct sum satisfies this definition?

Thanks for any help (sorry if my question is stupid or muddled)

share|improve this question
2  
That coproducts are unique up to unique isomorphism is easier to prove directly from the definition than in specific cases. –  Tobias Kildetoft Oct 5 '13 at 18:37
    
@TobiasKildetoft Yes, I have done this for products but not coproducts I was just looking at this specific example to try to understand and then getting bogged down. I will look at how to prove this "categorically" Thanks (As I imagine you can tell I have just started looking at category theory so still getting used to it. Thanks –  hmmmm Oct 5 '13 at 18:41
add comment

2 Answers 2

up vote 3 down vote accepted

Let's prove that the coproduct is unique in a category. Assume that there are two, let's say $Q$ and $Q'$ for the collection $\{A_a\}$. Let $\pi_a:A_a\to Q$ and $\pi':A_a\to Q'$ be the morphisms you talked about. Then by the universal property you mentioned, we have that there are morphisms $\phi:Q'\to Q$ and $\psi:Q\to Q'$ such that $\pi_a=\phi\circ\pi_a'$ and $\pi_a'=\psi\circ\pi_a$. Now these morphisms $\phi$ and $\psi$ are unique. If we now take $\phi\circ\psi:Q\to Q$, this also satisfies $\phi\circ\psi\circ\pi_a=\phi\circ\pi_a'=\pi_a$, the same as the identity. Since the map of the universal property is unique, we have that $\phi\circ\psi=\mbox{id}$. The same argument shows that $\psi\circ\phi=\mbox{id}$.

So the coproduct is unique. Therefore, if you can show that the direct sum of abelian groups satisfies what you are looking for (it seems you already did this?), then it must be "the" coproduct (up to isomorphism).

share|improve this answer
    
Thanks, yes from the comment I had basically done this (as it is the same as the proof that products are unique which I have). On a side not does the symbol $\exists !$ not mean there is one and only one? –  hmmmm Oct 5 '13 at 18:47
    
Yes I'm sorry! I didn't notice you had put that. My apologies. –  Robert Auffarth Oct 5 '13 at 19:02
    
Not a problem :) Thanks for the answer! –  hmmmm Oct 5 '13 at 19:07
add comment

I feel like this is one of those cases where the general proof can be a lot more enlightening than a special case, so here goes:

First, we need a few definitions. An object $I$ is a category is called initial if for any object $C$ in the category, there is a unique arrow $I\to C$.

Dually, an object $T$ is called terminal if for any object $C$ there is a unique arrow $C\to T$.

First thing we will now note is that if $I$ and $I'$ are initial objects of some category, then there is a unique arrow between them, and this arrow is an isomorphism. So initial objects are unique up to a unique isomorphism. The same is true for terminal objects.

Now, we need a bit more to be able to apply the above observations to cases like coproducts, since these are not initial or terminal in the category itself.

I will start with a special case of the construction, to make the notation simpler. Then, hopefully the more general version needed for product and coproduct will not seem too strange.

Let $D$ be some fixed object in the category. We will construct the category of "objects above $D$" as follows: The objects in this new category will be all pairs $(C,f)$ where $C$ is an object of the original category, and $f: C\to D$ is an arrow in the original category. Given two objects $(C,f)$ and $(C',f')$ in this new category, the arrows from $(C,f)$ to $(C',f')$ will be those arrows $g: C\to C'$ in the original category, such that $f'\circ g = f$. A good way to visualize this is to put the object $D$ at the bottom, and have the other objects form a line above it with their arrows pointing to $D$. Then an arrow between two of the objects is one that makes the triangle formed commute (I am not that familiar with how to make commutative diagrams here, but if there is any interest in it, I will give it a shot).

Now note that if $(I,f)$ is an initial object in this new category, and $(I',f')$ is another initial object, then these two are isomorphic via a unique isomorphism in the new category. But more importantly, they are also isomorphic (though the isomorphism need not quite be unique) in the original category. So it would make sense to define an object as being "the initial object above $D$", since this will give an object unique up to isomorphism (and if we include the information of the morphism to $D$, we get uniqueness of this isomorphism when we require that maps commute with those to $D$). Note, however, that the above definition only really makes sense if the new category actually has an initial object, which might not be the case.

Similarly, we can define the category of "objects below $D$", and talk of both initial and terminal above and below $D$, in all cases getting objects unique up to isomorphism when they exist.

Now to return to the case of coproduct, we need to generalize a bit, but not much. What we need to do is define the category of objects above or below a collection of objects. This is done completely analogously to the one for just one object, though we need to require several equalities like the $f'\circ g = f$ above (or the dual version when we take objects below instead of above), one for each object in the collection we start with.

Now it is just a matter of checking that the coproduct of a collection is the initial object below that collection (and the product is the terminal object above the collection).

The above construction of new categories can be even further generalized, but I will not go into the details here. A phrase to look up is "comma-categories", which is for example explained in Mac Lane's "Categories for the Working Mathematician" (I think he uses the opposite convention of what is above and what is below to what I do, and I think he uses the words under and over instead).

share|improve this answer
add comment

Your Answer

 
discard

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.