Deck transformation group of a regular covering as quotient
Let
Proof
We will show that an isomorphism is given by
where
is a path from to and . If
is an alternative path from to then and thus so
is independent of the choice of . It is also clear that
is a homomorphism, since and for any Let
such that . Then and thus (by First lemma Uniqueness). Then , so since the deck transformation group acts properly discontinuously . Therefore is a group monomorphism. Let
and let be the lift of with . Since acts transitively on fibres there exists a with , and thus . Therefore is a group epimorphism and thus an isomorphism.
In particular, if
Footnotes
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2010, Algebraische Topologie, ¶2.3.39, p. 97 ↩