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.