Covers of a space with cyclic fundamental group are determined by their degree #
A pointed connected cover of (X, x) is determined up to isomorphism by the subgroup of
π₁(X, x) it recovers (IsCoveringMap.exists_homeomorph_comp_eq_of_range_eq), and the index of
that subgroup is the number of sheets (IsCoveringMap.card_fiber_eq_index). In a cyclic group a
subgroup is determined by its index (IsCyclic.subgroup_eq_iff_index_eq). So when π₁(X, x) is
cyclic, two connected covers of X with the same number of sheets over x are isomorphic over
X, by an isomorphism carrying any chosen point of the first fibre to any chosen point of the
second.
The number of sheets is read with Nat.card, so an infinite fibre counts as 0; the statement
holds in that case as well, since in an infinite cyclic group the trivial subgroup is the only one
of index 0.
The main application is to the punctured disc, whose fundamental group is infinite cyclic: its
connected covers of finite degree e ≠ 0 are all isomorphic to z ↦ z ^ e.
Main declarations #
IsCoveringMap.exists_homeomorph_comp_eq_of_card_fiber_eq: over a base with cyclicπ₁(X, x), two pointed connected covers with fibres overxof the same cardinality are isomorphic as pointed covers.
References #
- A. Hatcher, Algebraic Topology, Cambridge University Press, 2002, Proposition 1.32 (the number of sheets is the index of the recovered subgroup) and Proposition 1.37 (pointed covers are classified by the recovered subgroup).
Over a base with cyclic fundamental group, a pointed connected cover is determined by its
number of sheets. If π₁(X, x) is cyclic and the fibres of p and q over x have the same
cardinality, then there is a homeomorphism E ≃ₜ F over X carrying the chosen point e₀ of
the first fibre to the chosen point f₀ of the second.