The set $5^{-\infty}\mathbb{Z}$ is a colimit I am trying to understand why the set of rational numbers whose denominators are powers of $5$, $5^{-\infty}\mathbb{Z}$, is a colimit. Specifically, why is $5^{-\infty}\mathbb{Z}$ the colimit of the diagram $$\mathbb{Z}\longrightarrow5^{-1}\mathbb{Z}\longrightarrow5^{-2}\mathbb{Z}\longrightarrow\cdots$$
I'm not sure what the maps to $5^{-\infty}\mathbb{Z}$ are. They can't be inclusion, or else we won't get commutativity. What am I missing?
I imagine the arrows above are given by division by $5$, and maybe the maps to the limit are multiplication by $5$?
 A: Hint: $5^{-n}\Bbb{Z}$ here means $\{5^{-n} x \mid x \in \Bbb{Z}\} \subseteq \Bbb{Q}$. The arrow $5^{-n}\Bbb{Z} \to 5^{-n-1}\Bbb{Z}$ is the inclusion function $x \mapsto x$ (a multiple of $1/5$ is also a multiple of $1/25$ etc.). The colimit construction gives that the elements  of the colimit are represented by finite sequences $(x_1, \ldots x_n)$ of integers with $x_n \neq 0$  (or $n = 1$ and $x_n = 0$) under an equivalence relations that means each such sequence can be identified with $x_n/5^n$.
A: $\require{AMScd}$First note that the colimit of the diagram (of sets or abelin groups)
$$\Bbb Z\xrightarrow{x\mapsto x/5}5^{-1}\Bbb Z\xrightarrow{x\mapsto x/5}5^{-2}\Bbb Z\to\cdots$$
is $\Bbb Z$ because we have a commutative diagram
\begin{CD}
\Bbb Z@>x\mapsto x/5>>5^{-1}\Bbb Z@>x\mapsto x/5>>5^{-2}\Bbb Z@>>>\cdots\\
@|@Vx\mapsto 5xVV@Vx\mapsto 5^2xVV\\
\Bbb Z@=\Bbb Z@=\Bbb Z@=\cdots
\end{CD}
where each arrow is an isomorphism (of sets or abelian groups).
On the other hand, the colimit of the diagram
$$\Bbb Z\hookrightarrow5^{-1}\Bbb Z\hookrightarrow 5^{-2}\Bbb Z\hookrightarrow\cdots$$
is
$$5^{-\infty}\Bbb Z=\bigcup_{n\in\Bbb N}5^{-n}\Bbb Z\tag 1$$
because if $M$ is an abelian group and $\varphi_n:5^{-n}\Bbb Z\to M$ for $n\in\Bbb N$ is a cocone of homomorphisms, then $\varphi_n|5^{-m}\Bbb Z=\varphi_m$ for every $m<n$.
By $(1)$, the map
\begin{align}
&5^{-n}x\mapsto\varphi_n(5^{-n}x)&&(x\in\Bbb Z)
\end{align}
defines the required homomorphism $\varphi:5^{-\infty}\Bbb Z\to M$.
