TITLE: What is the fibered coproduct of abelian groups?
QUESTION [5 upvotes]: From wikipedia:

I am not sure I understand this. For example:
Suppose we have $\alpha: \mathbb Z/(3) \to \mathbb Z/(6)$ where
$$0\mapsto 0, \quad 1 \mapsto 2, \quad 2 \mapsto 4,$$
and we have $\beta: \mathbb Z/(3) \to \mathbb Z/(9)$ where
$$0\mapsto 0, \quad 1 \mapsto 3, \quad 2 \mapsto 6.$$
Then we want the subgroup of $\mathbb Z/(6) \oplus \mathbb Z/(9)$ consisting of $(0,0), (2,-3), (4, -6)$? Why?
Why $(\alpha(z), -\beta(z))$? Why the negative $\beta(z)$?
How do we know the pairs consisting of $(f(z), -g(z))$ will always be a subgroup?

REPLY [3 votes]: First question:
The aim is to define a group $G$ such that the  following diagram is commutative:
\begin{alignat}{3}
\mathbf Z&/3\mathbf Z& &\xrightarrow{\enspace\beta\enspace} \mathbf Z&&/9\mathbf Z \\
&\llap{\alpha}\downarrow&&&&\downarrow \\
\mathbf Z&/6\mathbf Z& &\xrightarrow{\quad\quad}&&G
\end{alignat}
and $G$ has the universal property w.r.t. this diagram. So one first  defines maps  from $\mathbf Z/6\mathbf Z$ and $\mathbf Z/9\mathbf Z$ to their direct sum:
$$z\bmod 6\longmapsto(z\bmod 6,0),\qquad z\bmod 9\longmapsto(0,z\bmod 9).$$
Writing that the images of $\alpha(z)$ and $\beta(z)$ in a quotient of the direct sum are equal, one obtains the condition that the kernel from the direct sum to $G$ contains  all elements of the form $\;\bigl(\alpha(z),-\beta(z)\bigr)$.
Actually, the set of all these elements  is a subgroup of the direct sum, as results from $\alpha$ and $\beta$ being group homomorphisms.