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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 21G / Solution

Choose a path

γ

in

X

from

x_{0}

to

x_{1}

. For every

x \in p^{- 1} (x_{0})

, the path lifting theorem gives a unique lift of

γ

starting at

x

. Send

x

to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path lifting

p^{- 1} (x_{0}) ≅ p^{- 1} (x_{1}) .

In particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of

x_{0}

above

x_{0}

,

p_{*} π_{1} (X, x_{0}) ◃ π_{1} (X, x_{0}) .

The relevant lifting criterion for a covering space says that a based map

f : (Y, y_{0}) \to (X, x_{0})

lifts through

p

exactly when

f_{*} π_{1} (Y, y_{0}) \subseteq p_{*} π_{1} (X, x_{0}) .

For a universal covering,

X

is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface

Σ_{g}

. By the classification of connected covering spaces, degree-

n

connected covers correspond to index-

n

subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

$n = 1$ , because the subgroup is the whole fundamental group.
$n = 2$ , because every index-two subgroup is normal.
$g = 1$ , because $π_{1} (Σ_{1}) ≅ Z^{2}$ is abelian, so all its subgroups are normal.
For $g = 0$ , the sphere is simply connected, so a connected cover necessarily has $n = 1$ .

It remains to show that these are the only forced cases. Let

g \geq 2

and

n \geq 3

. Write

π_{1} (Σ_{g}) = ⟨ a_{1}, b_{1}, \dots, a_{g}, b_{g} \prod_{i = 1}^{g} [a_{i}, b_{i}] = 1 ⟩ .

In the symmetric group

S_{n}

, put

σ = (12 \dots n), τ = (12) .

Define a homomorphism by

a_{1} \mapsto σ, b_{1} \mapsto τ, a_{2} \mapsto τ, b_{2} \mapsto σ,

and send all remaining generators to the identity. The relation is respected because

[σ, τ] [τ, σ] = 1.

The cycle

σ

and transposition

τ

generate

S_{n}

, so the homomorphism is surjective.

Let

H = ϕ^{- 1} (Stab_{S_{n}} (1)) .

The natural action of

S_{n}

is transitive, so

H

has index

n

. Its image is the point stabilizer

S_{n - 1}

, which is not normal in

S_{n}

for

n \geq 3

; hence

H

is not normal. The connected covering corresponding to

H

is therefore an explicit degree-

n

nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly when

n \leq 2 or g \leq 1,

with the qualification that

g = 0

permits only

n = 1

, as summarized by the forced normality of finite connected covers of orientable surfaces.

Solved by gpt-5.6-sol high.

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