Deck transformation group of a regular covering as quotient

Let $𝑝 : ˜ 𝑋 ↠ 𝑋$ be a connected and path-connected regular covering, $𝑥 0 \in 𝑋$ , and $˜ 𝑥 0 \in 𝑝 - 1 {𝑥 0}$ . Let $𝐻 = 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ be the basepoint-invariant characteristic subgroup and $Γ = A u t 𝖢 𝗈 𝗏 𝑋 (𝑝)$ be the deck transformation group of $𝑝$ . Then¹ #m/thm/homotopy

Γ ≅ 𝜋 1 (𝑋, 𝑥 0) / 𝐻

Proof

We will show that an isomorphism is given by

Φ : Γ \to 𝜋 1 (𝑋, 𝑥 0) / 𝐻 𝛾 \mapsto [𝛼 𝛾] 𝐻 = 𝐻 [𝛼 𝛾]

where $˜ 𝛼 𝛾 : 𝕀 \to ˜ 𝑋$ is a path from $˜ 𝑥 0$ to $𝛾 (˜ 𝑥 0)$ and $𝛼 𝛾 = 𝑝 \circ ˜ 𝛼 𝛾$ .

If $˜ 𝛽 𝛾 : 𝕀 \to ˜ 𝑋$ is an alternative path from $˜ 𝑥 0$ to $𝛾 (˜ 𝑥 0)$ then $𝜋 1 𝑝 [――― ˜ 𝛽 𝛾 ⊙ ˜ 𝛼 𝛾] = [――― 𝛽 𝛾 ⊙ 𝛼 𝛾] \in 𝐻$ and thus

[𝛽 𝛾] 𝐻 = [𝛽 𝛾] [――― 𝛽 𝛾 ⊙ 𝛼 𝛾] 𝐻 = [𝛼 𝛾] 𝐻

so $Φ [𝛾]$ is independent of the choice of $𝛼 𝛾$ .

It is also clear that $Φ$ is a homomorphism, since $Φ (i d) = 𝐻$ and for any $𝛾, 𝜂 \in Γ$

Φ (𝛾 𝜂) = [𝛼 𝛾 𝜂] 𝐻 = [𝑝 \circ 𝛾 \circ ˜ 𝛼 𝜂] [𝛼 𝛾] = [𝛼 𝜂] [𝛼 𝛾] 𝐻 = Φ (𝛾) Φ (𝜂)

Let $𝛾 \in Γ$ such that $Φ (𝛾) = 𝐻$ . Then $[𝛼 𝛾] \in 𝐻$ and thus $[˜ 𝛼 𝛾] \in 𝜋 1 (˜ 𝑋, ˜ 𝑥 0)$ (by First lemma Uniqueness). Then $𝛾 (˜ 𝑥 0) = ˜ 𝛼 𝛾 (1) = ˜ 𝑥 0$ , so since the deck transformation group acts properly discontinuously $𝛾 = 𝑒$ . Therefore $Φ$ is a group monomorphism.

Let $[𝛽] 𝐻 \in 𝐺 / 𝐻$ and let $˜ 𝛽$ be the lift of $𝛽$ with $˜ 𝛽 (0) = ˜ 𝑥 0$ . Since $Γ$ acts transitively on fibres there exists a $𝛾 \in Γ$ with $𝛾 (˜ 𝑥 0) = ˜ 𝛽 (1)$ , and thus $Φ (𝛾) = [𝛽] 𝐻$ . Therefore $Φ$ is a group epimorphism and thus an isomorphism.

In particular, if $˜ 𝑋$ is simply connected then $Γ ≅ 𝜋 1 (𝑋, 𝑥 0)$ — see Universal covering.

#state/tidy | #lang/en | #SemBr

2010, Algebraische Topologie, ¶2.3.39, p. 97 ↩

Deck transformation group of a regular covering as quotient

Footnotes