Characteristic conjugacy class of a path-connected covering

Let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ be a path-connected covering with characteristic subgroup $𝐻 = i m 𝜋 1 𝑝$ . Then a subgroup $𝐻' \subseteq 𝜋 1 (𝑋, 𝑥 0)$ is conjugate to $𝐻$ iff it is the characteristic subgroup of $𝑝$ with respect to a different basepoint $˜ 𝑥' 0$ . #m/thm/homotopy

Proof

For the reverse direction, let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ and $𝑝' : (˜ 𝑋, ˜ 𝑥' 0) ↠ (𝑋, 𝑥 0)$ be the same path-connected covering considered with different basepoints, with characteristic subgroups $𝐻$ and $𝐻'$ respectively Let $˜ 𝛾 : 𝕀 \to ˜ 𝑋$ be a path from $˜ 𝑥 0$ to $˜ 𝑥' 0$ . We can define an isomorphism

Φ : 𝜋 1 (˜ 𝑋, ˜ 𝑥 0) \to 𝜋 1 (˜ 𝑋, ˜ 𝑥' 0) [˜ 𝛼] \mapsto [˜ 𝛾 ⊙ ˜ 𝛼 ⊙ ―― ˜ 𝛾]

Then

𝐻' = 𝜋 1 𝑝' (𝜋 1 (˜ 𝑋, ˜ 𝑥' 0)) = 𝜋 1 𝑝' (Φ (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))) = [𝛼] ⊙ 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0)) ⊙ [𝛼] - 1 = [𝛼] ⊙ 𝐻 ⊙ [𝛼] - 1

hence the characteristic groups are conjugate.

For the forward direction, let $𝐻 = 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ and $𝐻' = [𝛼] ⊙ 𝐻 ⊙ [𝛼] - 1$ for some closed loop $𝛼 : 𝕀 \to 𝐻$ . Then $𝛼$ has a unique lift with $˜ 𝛼 (0) = ˜ 𝑥 0$ . Then $𝐻'$ is the characteristic group of $𝑝$ with basepoint $˜ 𝑥' 0$ .

Hence a covering without choice of basepoint corresponds to a conjugation class of subgroups of $𝜋 1 (𝑋, 𝑥 0)$ .

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