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A homomorphism induces a continuous homomorphism of profinite completions

Statement

A group homomorphism induces a continuous homomorphism of profinite completions, and these induced maps respect identities and compositions.

Facts & Assumptions

Given: A homomorphism $f : G \to H$ .

[L1]

The profinite completion satisfies the universal property with respect to continuous maps into profinite groups (The profinite completion is initial among continuous homomorphisms from G to profinite groups).

[F1]

The completion $\hat{H}$ is the inverse limit of the finite discrete quotients of $H$ , hence is a profinite topological group, and $ι_{H}$ is its canonical map (The profinite completion is the inverse limit of the finite quotients G over N, The canonical map sends g to its coherent system of residue classes).

[L2]

A map into an inverse limit is continuous exactly when all its coordinate composites are continuous (A map into an inverse limit is continuous exactly when all coordinate composites are continuous).

Proof

technique · direct

1.1

By [F1], $\hat{H}$ is a profinite topological group.

F1given

2.1

Fix a finite-index normal subgroup $N ⊴ H$ . The kernel of the coordinate map $G \overset{f}{\to} H \to H / N$ is $f^{- 1} (N)$ , and the induced map $G / f^{- 1} (N) \to H / N$ is injective. Thus $f^{- 1} (N)$ is finite-index normal in $G$ . Every fibre of the coordinate map is a coset of this open subgroup, so the coordinate map to the finite discrete group $H / N$ is continuous. Since this holds for every coordinate, [L2] makes $ι_{H} \circ f : G \to \hat{H}$ continuous for the profinite topology on $G$ .

F1L2step 1.1algebra

3.1

Apply [L1] to the continuous homomorphism from step 2.1. It gives a unique continuous homomorphism $\hat{f} : \hat{G} \to \hat{H}$ satisfying $\hat{f} \circ ι_{G} = ι_{H} \circ f$ .

L1step 1.1step 2.1construct

4.1

If $f = {id}_{G}$ , then both $\hat{f}$ and ${id}_{\hat{G}}$ compose with $ι_{G}$ to give $ι_{G}$ . Uniqueness in [L1] therefore forces $\hat{{id}_{G}} = {id}_{\hat{G}}$ .

L1step 3.1

5.1

For homomorphisms $G \overset{f}{\to} H \overset{g}{\to} K$ , both $\hat{g} \circ \hat{f}$ and $\hat{g \circ f}$ compose with $ι_{G}$ to give $ι_{K} \circ g \circ f$ . Uniqueness in [L1] again yields $\hat{g \circ f} = \hat{g} \circ \hat{f} .$ So profinite completion is functorial.

L1step 3.1algebra∎

Depends on

Used by

Nonisomorphic groups can share the same profinite completion Counterexample
FALSE: isomorphic profinite completions force the original groups to be isomorphic False statement

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Brian Osserman, Math 6112 notes on inverse limits and profinite groups (standard reference, not scraped)
H. W. Lenstra, Profinite groups and Galois groups (standard reference, not scraped)