Computing easy direct limit of groups How do I start computing easy direct limit of groups:
1) $\mathbb{Z} \overset{1}\longrightarrow \mathbb{Z} \overset{2}\longrightarrow \mathbb{Z} \overset{3}\longrightarrow \mathbb{Z} \overset{4}\longrightarrow\mathbb{Z}\longrightarrow\cdots$ where the maps are just multiplication by "above arrow" numbers. 
2) $\mathbb{Z}\overset{2}\longrightarrow \mathbb{Z} \overset{2}\longrightarrow \mathbb{Z} \overset{2}\longrightarrow \mathbb{Z} \overset{2}\longrightarrow$$\mathbb{Z}\longrightarrow\cdots $ where the maps are just multiplication by 2. 
How do I get intuition for the resulting direct limit? 
 A: Having worked only with inverse limits (of topological spaces), I was curious to see what I could make of one of these direct limits by working directly(!) from the definition. For the first one let $G_k=\Bbb Z$ for $k\in\Bbb Z^+$. For each $k\in\Bbb Z^+$ we have the bonding map $$f_{k,k+1}:G_k\to G_{k+1}:n\mapsto kn\;,$$ so we must have 
$$f_{k,\ell}:G_k\to G_\ell:n\mapsto k(k+1)(k+2)\ldots(\ell-1)n=k^{\overline{\ell-k}}n$$
for $k,\ell\in\Bbb Z^+$ with $k\le\ell$, where $k^{\overline{\ell-k}}$ is a rising factorial.
The next step is to form $\bigsqcup_{k\in\Bbb Z^+}G_k$, the disjoint union of the groups $G_k$. One way to do this is to replace $G_k$ by $\Bbb Z\times\{k\}$, defining addition in this new object by $\langle m,k\rangle+\langle n,k\rangle=\langle m+n,k\rangle$; in other words, the $k$ in the second component just comes along for the ride. We can then identify the disjoint union with the set $\Bbb Z\times\Bbb Z^+$, whose element $\langle n,k\rangle$ represents the element $n$ of $G_k$.
We now define an equivalence relation $\sim$ on this disjoint union: $\langle m,j\rangle\sim\langle n,k\rangle$ iff there is some $\ell\ge j,k$ such that $f_{j,\ell}(m)=f_{k,\ell}(n)$, i.e., such that $j^{\overline{\ell-j}}m=k^{\overline{\ell-k}}n$. Without loss of generality assume that $j\le k$. Then $\langle m,j\rangle\sim\langle n,k\rangle$ iff there is an $\ell\ge k$ such that 
$$f_{k,\ell}(n)=f_{j,\ell}(m)=f_{k,\ell}\big(f_{j,k}(m)\big)\;,$$
and $f_{k,\ell}$ is injective, so $n=f_{j,k}(m)$.
Thus, $$\langle n,k\rangle\sim\langle kn,k+1\rangle\sim\langle k(k+1)n,k+2\rangle\sim\langle k(k+1)(k+2)n,k+3\rangle\sim\ldots\;.$$ If $k-1\mid n$, we can chain back a step: $$\langle n,k\rangle\sim\left\langle\frac{n}{k-1},k-1\right\rangle\;;$$ otherwise we cannot, and $\langle n,k\rangle\not\sim\langle m,j\rangle$ for any $j<k$ and $m\in\Bbb Z$. It’s not hard now to verify the following proposition.

Proposition. For any $\langle n,k\rangle\in\Bbb Z\times\Bbb Z^+$ there is a unique $\langle m,j\rangle\in\Bbb Z\times\Bbb Z^+$ such that 
  
  
*
  
*$j\le k$;  
  
*$\langle m,j\rangle\sim\langle n,k\rangle$; and  
  
*$j-1\nmid m$.
  

The direct limit is now defined as the quotient $$G=\left(\bigsqcup_{k\in\Bbb Z^+}G_k\right)\Bigg/\sim\;.$$
With the aid of the proposition we can identify $G$ with $H=\{\langle n,k\rangle\in\Bbb Z\times\Bbb Z^+:k-1\nmid n\}$. The group operation $\oplus$ on $H$ can then be defined as follows: if $j\le k$, $\langle m,j\rangle\oplus\langle n,k\rangle$ is the unique $\langle r,\ell\rangle\in H$ such that 
$$\langle r,\ell\rangle\sim\left\langle f_{j,k}(m)+n,k\right\rangle\;.$$
Note that if $j<k$, then $k-1\mid f_{j,k}(m)$; $k-1\nmid n$, since $\langle n,k\rangle\in H$, so $k-1\nmid f_{j,k}(m)+n$, and $$\left\langle f_{j,k}(m)+n,k\right\rangle\in H\;.$$
For example, if $2\nmid m$ and $4\nmid n$, so that $\langle m,3\rangle,\langle n,5\rangle\in H$, then $$\langle m,3\rangle\oplus\langle n,5\rangle=\langle 3\cdot4\cdot m+n,5\rangle=\langle 12m+n,5\rangle\;.$$
More generally, if $\langle m,j\rangle,\langle n,k\rangle\in H$ with $j<k$, then
$$\langle m,j\rangle\oplus\langle n,k\rangle=\left\langle j^{\overline{k-j}}m+n,k\right\rangle\;,\tag{1}$$
where the first components look an awful lot like the numerators in the addition $\frac{m}a+\frac{n}b$ if $b=j^{\overline{k-j}}a$. The second components $j$ and $k$ should somehow correspond to the denominators $a$ and $b$, and a little tinkering reveals that the map $$\langle n,k\rangle\mapsto\frac{n}{(k-1)!}$$ turns $(1)$ into
$$\frac{m}{(j-1)!}+\frac{n}{(k-1)!}=\frac{j^{\overline{k-j}}m+n}{(k-1)!}$$
if we replace $\oplus$ by $+$. What happens when $j=k$? Then
$$\langle m,k\rangle\oplus\langle n,k\rangle\leadsto\langle m+n,k\rangle\;,$$
which may not be in $H$, and $$\frac{m}{(k-1)!}+\frac{n}{(k-1)!}=\frac{m+n}{(k-1)!}\;,$$ which may be reducible to a fraction with a smaller factorial denominator. Moreover, $\langle m+n,k\rangle\sim\langle r,j\rangle\in H$ iff $\frac{m+n}{(k-1)!}$ reduces to $\frac{r}{(j-1)!}$.
Thus, if we let $$F=\left\{\frac{m}{n!}:n\nmid m\right\}\;;$$ then $\langle H,\oplus\rangle\cong\langle F,+\rangle$. Finally, let $\frac{m}n\in\Bbb Q$, where $m,n\in\Bbb Z$, $n>0$, and $m$ and $n$ are relatively prime. Let $k$ be the smallest non-negative integer such that $n\mid k!$, and let $d=\frac{k!}n$; then $\frac{m}n=\frac{dm}{k!}$, which is equal to a unique member of $F$. Thus, $\langle F,+\rangle$ is nothing but $\langle\Bbb Q,+\rangle$.
A: I remarked in another question on a trick to reinterpret the objects and morphism of the system as taking place in a different domain so that the direct limit can be seen more "directly" as a union.
The system ${\Bbb Z}\xrightarrow{1}{\Bbb Z}\xrightarrow{2}{\Bbb Z}\xrightarrow{3}{\Bbb Z}\xrightarrow{4}{\Bbb Z}\xrightarrow{5}\cdots$ is equivalent to the system of embeddings
$${\Bbb Z}\hookrightarrow\frac{1}{1}{\Bbb Z}\hookrightarrow\frac{1}{2\cdot1}{\Bbb Z}\hookrightarrow \frac{1}{3\cdot2\cdot1}{\Bbb Z}\hookrightarrow\frac{1}{4\cdot3\cdot2\cdot1}{\Bbb Z}\hookrightarrow\cdots \subset {\Bbb Q}$$
The union is easily seen to be $\bigcup \frac{1}{n!}{\Bbb Z}=\Bbb Q$.
The system ${\Bbb Z}\xrightarrow{2}{\Bbb Z}\xrightarrow{2}{\Bbb Z}\xrightarrow{2}{\Bbb Z}\xrightarrow{2}{\Bbb Z}\xrightarrow{2}\cdots$ is equivalent to the system of embeddings
$${\Bbb Z}\hookrightarrow\frac{1}{2}{\Bbb Z}\hookrightarrow\frac{1}{2^2}{\Bbb Z}\hookrightarrow\frac{1}{2^3}{\Bbb Z}\hookrightarrow\frac{1}{2^4}{\Bbb Z}\hookrightarrow\cdots\subset{\Bbb Z}[1/2].$$
The union is easily seen to be $\bigcup\frac{1}{2^n}{\Bbb Z}={\Bbb Z}[1/2]$.
