Existence and Uniqueness of Direct Limit of Sequence of Groups

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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:

$\left({x_n, n}\right) \sim \left({y_m, m}\right) \iff \exists k \ge n,m: g_{n,k} \left({x_n}\right) = g_{m,k} \left({y_m}\right)$


Now define $G_\infty$ as the quotient set:

$G_\infty := \hat G_\infty / \sim$



Subsequently define $u_n: G_n \to G_\infty$ by:

$u_n \left({x_n}\right) := \left[\!\left[{\left({x_n, n}\right)}\right]\!\right]$



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.



$\Box$


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