I'm just learning the concept of direct limit. I did problem $8$a fromt Dummit Foote section $7.6$.
As an example, someone suggested I look $$\mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \xrightarrow{\times 3} \mathbb{Z}\xrightarrow{\times 4} \mathbb{Z}\xrightarrow{\times 5} \mathbb{Z}\xrightarrow{\times 6} \cdots$$ and I should get the rationals.
I've tried playing around a bit, seeing which elements are equivalent in the quotient of the disjoint union modulo the relation, but not sure how to get the rationals.
Any help on how to get the rationals out of this direct limit?
Here is a directed system of abelian groups which is isomorphic to yours: $$ \mathbb{Z} \to \frac{1}{2}\mathbb{Z} \to \frac{1}{6}\mathbb{Z} \to \ldots \to \frac{1}{n!} \mathbb{Z} \to \ldots $$ (here we mean the cyclic subgroup of $\mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $\mathbb{Q}$ as a limit. (The inclusion into $\mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $\mathbb{Q}$ is eventually in $\frac{1}{n!}\mathbb{Z}$ for some $n!$).