Deck transformation

A covering is regular iff its deck transformation group acts transitively on fibres

Let be a connected and locally path-connected covering and be its deck transformation group. Then is a regular covering iff acts on one (and therefore every) fibre transitively. 1 #m/def/homotopy Equivalently, the orbit of each is a whole fibre.

Proof

Let be a regular covering and . Let , and let and denote the characteristic subgroups with respect to and respectively. Since is regular, , and by equivalence of coverings criterion the coverings with either base point are equivalent. Hence there exists such that .

For the converse, assume acts transitively on , i.e. the following diagram commutes for any :

https://q.uiver.app/#q=WzAsNCxbMCwwLCIoXFx0aWxkZSBYLCBcXHRpbGRlIHhfMCkiXSxbNCwwLCIoXFx0aWxkZSBYLCBcXHRpbGRlIHhfMCcpIl0sWzIsMiwiKFgsIHhfMCkiXSxbNiwwXSxbMCwyLCJwIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMiwicCIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFswLDEsIlxcZ2FtbWEiLDAseyJjdXJ2ZSI6LTF9XSxbMSwwLCJcXGdhbW1hXnstMX0iLDAseyJjdXJ2ZSI6LTF9XV0=

Applying the Fundamental group functor to this diagram it is clear that the characteristic subgroups is basepoint-invariant. Therefore is a regular covering.

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Footnotes

  1. 2010, Algebraische Topologie, ¶2.3.36, p. 96 ↩