Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 1

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Lemma

On $\widehat G_\infty := \ds \coprod_{n \mathop \in \N} G_n$ the relation:

$\tuple {x_n, n} \sim \tuple {y_m, m} \iff \exists k \ge n, m: \map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$

is an equivalence relation.

Proof

Reflexivity

Since $g_{n, n} = \mathop {Id}_{G_n}$ we have:

$\forall \tuple {x_n, n} \in \widehat G_\infty: \map {g_{n, n} } {x_n} = \map {g_{n, n} } {x_n}$

Hence:

$\tuple {x_n, n} \sim \tuple {x_n, n}$

$\Box$

Symmetry

Let $\tuple {x_n, n} \sim \tuple {y_m, m}$.

Then there exists a $k \ge n, m$ such that:

$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {x_m}$

Hence also:

$\map {g_{m, k} } {x_m} = \map {g_{n, k} } {x_n}$

That is:

$\tuple {y_m, m} \sim \tuple {x_n, n}$

$\Box$

Transitivity

Let $\tuple {x_n, n} \sim \tuple {y_m, m}$ and $\tuple {y_m, m} \sim \tuple {z_r, r}$.

Then there exist $k \ge m, n$ and $l \ge n, r$ such that:

$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$

$\map {g_{m, l} } {y_m} = \map {g_{r, l} } {z_r}$

Let $q:= \max \set {k, l}$.

Then we have:

\(\ds \map {g_{n, q} } {x_n}\)

\(=\)

\(\ds \map {g_{k, q} } {\map {g_{m, k} } {y_m} }\)

\(\ds \)

\(=\)

\(\ds \map {g_{m, q} } {y_m}\)

\(\ds \)

\(=\)

\(\ds \map {g_{l,q} } {\map {g_{m, l} } {y_m} }\)

\(\ds \)

\(=\)

\(\ds \map {g_{l, q} } {\map {g_{r, l} } {z_r} }\)

\(\ds \)

\(=\)

\(\ds \map {g_{r,q} } {z_r}\)

that is:

$\tuple {x_n, n} \sim \tuple {z_r, r}$

$\blacksquare$