Existence and Uniqueness of Direct Limit of Sequence of Groups
Contents
Theorem
Let $\left({G_n}\right)_{n \in \N}$ be a sequence of groups.
Let $\left({g_n}\right)_{n \in \N}, g_n: G_n \to G_{n+1}$ be a sequence of group homomorphisms.
Then their direct limit $G_\infty$ exists and is unique up to unique isomorphism.
Proof
Existence
Define $\hat G_\infty$ by:
- $\hat G_\infty := \displaystyle \bigsqcup_{n \mathop \in \N} G_n$
where $\displaystyle \bigsqcup$ signifies a disjoint union.
For $m \ge n$, denote with $g_{n,m}$ the composition $g_{m-1} \circ \cdots \circ g_n: G_n \to G_m$.
In particular, $g_{n,n}: G_n \to G_n$ is taken to be the identity mapping on $G_n$.
Now define an equivalence relation $\sim$ on $\hat G_\infty$ by putting:
- $\left({x_n, n}\right) \sim \left({y_m, m}\right) \iff \exists k \ge n,m: g_{n,k} \left({x_n}\right) = g_{m,k} \left({y_m}\right)$
Put a lemma or so proving this is an equivalence
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Now define $G_\infty$ as the quotient set:
- $G_\infty := \hat G_\infty / \sim$
group operation on $G_\infty$: define and well-define
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Subsequently define $u_n: G_n \to G_\infty$ by:
- $u_n \left({x_n}\right) := \left[\!\left[{\left({x_n, n}\right)}\right]\!\right]$
prove $u_n$ is group hom
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Now let $H$ be a group and $h_n: G_n \to H$ be group homomorphisms such that $h_{n+1} \circ g_n = h_n$.
Define $h_\infty: G_\infty \to H$ by:
- $h_\infty \left({\left[\!\left[{\left({x_n, n}\right)}\right]\!\right]}\right) = h_n \left({x_n}\right)$
If we can show $h_\infty$ is well-defined and a group homomorphism, we are done, since under this definition:
- $h_\infty \circ u_n \left({x_n}\right) = h_n \left({x_n}\right)$
for all $x_n \in G_n$, so $h_\infty \circ u_n = h_n$ by Equality of Mappings.
prove $h_\infty$ well-def, grp hom
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$\Box$
Uniqueness
From Equivalence of Definitions of Direct Limit of Sequence of Groups, $G_\infty$ can be described as a colimit in the category of groups $\mathbf{Grp}$.
By Colimit is Unique, it therefore is unique up to unique $\mathbf{Grp}$-isomorphism.
By Isomorphism in Category of Groups iff Group Isomorphism, the result follows.
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 5.6$: Example $5.29$
