I have seen that the following theorem, known as universal property of the direct sum, is always stated for direct sums of abelian groups:
(Universal mapping property of external direct sum): let
$\{G_{s} | s \in S\}$ be a nonempty set of abelian groups, and let $\bigoplus_{s\in S}G_{s}$ be the external direct sum, the associated group homomorphisms being the embedding mappings $i_{s_0} : G_{s_0} → \bigoplus_{s\in S}G_{s}$. If $H$ is any abelian group and $\{\phi_{s} ~\mid~ s \in S\}$ is a system of group homomorphisms $\phi_{s} : G_{s} → H$, then there exists a unique group homomorphism $\phi :\bigoplus_{s\in S}G_{s} → H$ such that $\phi \circ i_{s_0} = \phi_{s_0}$ for all $s_{0}\in S$.
What I do not get is the following: while on one hand the proof of the theorem looks to me not to depend on the commutativity of the family of groups (but I might have missed it), on the other hand it is always stated for direct sums of abelian groups.
Is there an example or a general principle showing what could go wrong if we drop the hypothesis?
Have a nice day!
Best Answer
The apparent issue comes from your statement of the Universal Property, and by clarifying that statement, we will see that the issue resolves itself.
In a fixed category $\mathcal{C}$, we define the Coproduct of two objects $A$ and $B$ to be
satisfying the following Universal Property:
Now, in the category of Abelian Groups (Ab), one can check that $A \oplus B$ is the coproduct of $A$ and $B$ (interestingly, it is also the product). However, the direct product of two groups does not satisfy the universal property of the coproduct. Indeed, that is why we don't call it a direct sum anymore.
I encourage you to work out the following example to see why this isn't the case:
The issue, I'm sure you'll find, is that $(1,0)$ and $(0,1)$ commute in $\mathbb{Z} \oplus \mathbb{Z}$, but their images under $\varphi$ and $\psi$ don't commute in $\mathfrak{S}_3$.
But the universal property as you stated in your question restricts the codomain to be abelian groups, which completely sidesteps this problem. If we are only working with abelian groups, this is great! But in the category of all groups, such arbitrary restrictions on the codomain aren't allowed.
Hope is not lost, though! There is a coproduct in the category of all groups. It is called the free product, and is extremely interesting, though I'm afraid this answer is already getting a bit long.
As a parting word, there is also categorical justification for your observation that, when restricting the codomain to abelian groups, the direct sum approach stil works. Using only group theory, one could verify that the direct sum of abelianizations of groups is the same thing as the abelianization of their free product. At a slightly higher level of abstraction, the abelianization "functor" is a left adjoint. As such, it preserves coproducts, as you have noticed.
I hope this helps ^_^