Viewing the rational numbers as a direct limit

Question

I'm trying to understand why the rational numbers can be interpreted as the following colimit:

$\mathbb{Q} = \varinjlim \frac{1}{n}\mathbb{Z}$ (Vakil, Rising Sea, Ex. 1.4C)

I'm not looking for a solution, but more so looking for help with the following two parts of this:

1. What the maps in the $\frac{1}{n} \mathbb{Z}$ diagram are: I see that if $n | m$, say $n \cdot d = m$, then the map $\frac{1}{n}\mathbb{Z} \rightarrow \frac{1}{m}\mathbb{Z}$ is the map $\frac{a}{n} \mapsto \frac{da}{dn} = \frac{da}{m}$. But what do we do about the maps from $\frac{1}{n}\mathbb{Z} \rightarrow \frac{1}{n+1}\mathbb{Z}$ look like? My first guess was $\frac{a}{n} \mapsto \frac{(n+1)a}{(n+1)n}$ but this doesn't commute (or rather, the composition of the maps $\frac{1}{n}\mathbb{Z} \rightarrow \frac{1}{n+1}\mathbb{Z} \rightarrow ... \rightarrow \frac{1}{m}\mathbb{Z}$) with the map above when n|m as far as I can see. Is this map correct and I'm just missing something or is there a different map?

2. My guess for the maps $\frac{1}{n}\mathbb{Z} \rightarrow \mathbb{Q}$ as inclusion. Is this correct? Given that the way I am thinking about the problem, the maps between the $\frac{1}{n}\mathbb{Z}s$ do not really change the fraction in the rationals so the relevant diagrams should commute.

Thanks

Edit: I have indeed seen these stack exchange posts Here and Here, which gave me the guesses as to what the maps should be. I'm more so looking to see if the guess is correct and what I'm missing with the commutativity of the maps in my first question (if the map is fine).

Answer 1

Your guesses are correct that all the maps are just inclusion maps. What you are missing is that the index category of the colimit is not the poset of positive integers ordered with their usual order but rather the poset of positive integers ordered by divisibility. That is, it is the category where the objects are positive integers and there is a unique morphism $m\to n$ whenever $m$ divides $n$. This means you only need a morphism $\frac{1}{m}\mathbb{Z}\to\frac{1}{n}\mathbb{Z}$ when $m$ divides $n$, not a morphism $\frac{1}{n}\mathbb{Z}\to\frac{1}{n+1}\mathbb{Z}$ for each $n$.