Main theorem of coverings

Let $(𝑋, 𝑥 0)$ be a locally path-connected, connected, and semilocally simply connected topological space. Then for every subgroup $𝐻 \subseteq 𝜋 1 (𝑋, 𝑥 0)$ there exists a covering $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ unique up to equivalence with characteristic subgroup $𝐻$ . #m/thm/homotopy

Construction

Take the universal covering $ˆ 𝑝 : ˆ 𝑋 ↠ 𝑋$ and consider $Φ (𝐻) \subseteq Γ$ where $Φ : 𝜋 1 (˜ 𝑋, 𝑥 0) \to Γ$ is an isomorphism. The covering is given by $˜ 𝑋 = ˆ 𝑋 / Φ (𝐻)$ with

𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0) ˆ 𝑥 Φ (𝐻) \mapsto ˆ 𝑝 (ˆ 𝑥 0)

Proof

Uniqueness up to equivalence follows from equivalence of coverings criterion. Since $(𝑋, 𝑥 0)$ is semilocally simply connected, it has a universal covering $ˆ 𝑝 : (ˆ 𝑋, ˆ 𝑥 0) ↠ (𝑋, 𝑥 0)$ . Let $Γ = A u t 𝖢 𝗈 𝗏 𝑋 (ˆ 𝑝)$ According to Deck transformation group of a regular covering as quotient

Φ : 𝜋 1 (𝑋, 𝑥 0) \to Γ [𝛼] \mapsto (ˆ 𝑥 0 \mapsto ˆ 𝛼 (1))

is an isomorphism, where $ˆ 𝛼$ is the unique lift of $𝛼$ with $ˆ 𝛼 (0) = ˆ 𝑥 0$ , and $(ˆ 𝑥 0 \mapsto ˆ 𝛼 (1))$ denotes a unique deck transformation with this property.

Now take the orbit space $˜ 𝑋 = ˆ 𝑋 / Φ (𝐻)$ with the canonical projection

𝑓 : (ˆ 𝑋, ˆ 𝑥 0) ↠ (˜ 𝑋, ˜ 𝑥 0) ˆ 𝑥 \mapsto Φ (𝐻) ˆ 𝑥

Since the deck transformation group acts properly discontinuously, so too does $Φ (𝐻) \subseteq Γ$ , and the orbit space of a properly discontinuous effective group action forms a covering, which in this case is universal. Thus

A u t 𝖢 𝗈 𝗏 ˜ 𝑋 (𝑓) ≅ 𝜋 1 (˜ 𝑋, ˜ 𝑥 0) ≅ Φ (𝐻) ≅ 𝐻

We now define

𝑝 : (˜ 𝑋, ˜ 𝑥 0) \to (𝑋, 𝑥 0) 𝑓 (ˆ 𝑥) \mapsto ˆ 𝑝 (ˆ 𝑥)

which is well-defined since $𝑓 (ˆ 𝑥) = 𝑓 (ˆ 𝑥')$ iff $ˆ 𝑥' = 𝛾 (ˆ 𝑥)$ for some $𝛾 \in 𝐻 \subseteq Γ$ , and then $ˆ 𝑝 \circ 𝛾 (ˆ 𝑥) = ˆ 𝑝 (ˆ 𝑥)$ ; and continuous by Universal property.

Now let $𝑥 \in 𝑋$ and let $𝑈$ be a neighbourhood of $𝑥$ evenly covered by $ˆ 𝑝$ with sheets ${ˆ 𝑈 𝑖} 𝑖 \in 𝐼$ . Let $𝐽 \subseteq 𝐼$ such that for all $𝑖 \in 𝐼$ there exists exactly one $𝑗 \in 𝐽$ such that $𝑓 (ˆ 𝑈 𝑖) = 𝑓 (ˆ 𝑈 𝑗)$ , and let $˜ 𝑈 𝑗 = 𝑓 (ˆ 𝑈 𝑗)$ . Then

𝑝 - 1 (𝑈) = ∐ 𝑗 \in 𝐽 ˜ 𝑈 𝑗

and $(𝑝 ↾ ˜ 𝑈 𝑗) - 1 = (𝑓 ↾ ˆ 𝑈 𝑗) \circ (ˆ 𝑝 ↾ ˆ 𝑈 𝑗) - 1$ , therefore $𝑝$ is a covering. Then by construction

[𝛼] \in 𝐻 ⟺ (ˆ 𝑥 0 \mapsto ˆ 𝛼 (1)) \in Φ (𝐻) ⟺ ˜ 𝛼 (1) = 𝑓 \circ ˆ 𝛼 (1) = ˜ 𝑥 0 ⟺ [𝛼] = 𝜋 1 𝑝 [˜ 𝛼]

so $𝐻 = 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ as required.

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