Equivalence of coverings criterion

Let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ and $𝑞 : (˜ 𝑋', ˜ 𝑥' 0) ↠ (𝑋, 𝑥 0)$ be connected and locally path-connected coverings. Then $𝑝$ and $𝑞$ are equivalent iff $i m 𝜋 1 𝑝 = i m 𝜋 1 𝑞$ , #m/thm/homotopy i.e. iff they have the same characteristic subgroup.

Proof

By discussion in Category of coverings with basepoint, if $i m 𝜋 1 𝑝 = i m 𝜋 1 𝑞$ there exists a unique $𝑓 \in 𝖢 𝗈 𝗏 (𝑋, 𝑥 0) (𝑝, 𝑞)$ and $𝑔 \in 𝖢 𝗈 𝗏 (𝑋, 𝑥 0) (𝑞, 𝑝)$ . Moreover, the identities $i d 𝑝 = i d (˜ 𝑋, ˜ 𝑥 0)$ and $i d 𝑞 = i d (˜ 𝑋', ˜ 𝑥' 0)$ are the only morphisms in $𝖢 𝗈 𝗏 (𝑋, 𝑥 0) (𝑝, 𝑝)$ and $𝖢 𝗈 𝗏 (𝑋, 𝑥 0) (𝑞, 𝑞)$ respectively. Therefore $𝑓 𝑔 = i d 𝑝$ and $𝑔 𝑓 = i d 𝑞$ , hence $𝑝$ and $𝑞$ are equivalent.

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