What is the fibered coproduct of abelian groups?

Question

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?

Answer 1

This term 'gluing' probably comes from topology.

If $f:A\to B$ and $g:A\to C$, then we want to consider $B+C$ (this is just the disjoint union in the category of topological spaces) and glue $B$ and $C$ along $f$ and $g$, by identifying $f(a)$ with $g(a)$ for each $a\in A$.
That is, we take the quotient by the equivalence relation generated by the relation $f(a)\sim g(a)$, so that in the quotient we will effectively have $f(a)=g(a)$.

The same is happening for Abelian groups.
The coproduct is the direct sum, and we want to enforce $f(a)=g(a)$, or equivalently, $f(a)-g(a)=0$, for all $a\in A$, so we take the subgroup generated by the elements $f(a)-g(a)$ in $B\oplus C$, and take the quotient by this.

Also note that in your example, we have $-(2,-3)=(-2,3)=(4,-6)$.

Answer 2

Gluing is maybe a bit misleading in some categories, but it basically just means identifying some stuff. What you want to identify are the images of the given maps (here $\alpha$ and $\beta$) inside the coproduct (here the direct sum). If you do this in some more geometric categories like the category of topological spaces you can also see where the term gluing comes from. In our case we want to ensure that the equation $\alpha(g) = \beta(g)$ holds inside the direct sum, which is equivalent to forcing the difference $\alpha(g) - \beta(g) = 0$. Therefore we get the sign that you were wondering about. Making sure that one of these equivalent equations hold means taking the quotient by the subgroup that is defined by the equation. This is a very natural setting that we use quite a lot. For example if you want to make a group abelian you quotient out the commutators to force them to be trivial and thus get the abelianization.

And towards your other question: You can just do the computation to see that these tuples form a group.

If you want to understand pushouts more from an abstract point of view then try to construct them via coproducts and coequalizers. As coequalizers are the objects that make morphisms coincide, this is the way to go.

Answer 3

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.