A covering is injective on the fundamental group

Let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ be a covering. Then the induced homomorphism $𝜋 1 𝑝 : 𝜋 1 (˜ 𝑋, ˜ 𝑥 0) ↣ 𝜋 1 (𝑋, 𝑥 0)$ is a group monomorphism. #m/thm/homotopy

Proof

Let $[˜ 𝛼] \in k e r 𝜋 1 𝑝$ . Then $𝛼 = 𝑝 \circ ˜ 𝛼 ≃ 𝑐 𝑥 0$ . But $˜ 𝛼$ and $𝑐 ˜ 𝑥 0$ are the lifts of $𝛼$ and $𝑐 𝑥 0$ respectively, so by Third lemma Lifts of homotopies of paths $˜ 𝛼 ≃ 𝑐 ˜ 𝑥 0$ , Therefore $k e r 𝜋 1 𝑝 = {𝑒}$ and thus $𝑝$ is a Group monomorphism.

Therefore the characteristic subgroup of the covering $𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ is isomorphic to $𝜋 1 (˜ 𝑋, ˜ 𝑥 0)$ .

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