It's a known fact (see, for example, "Algebraic Topology" by Hatcher, chapters) $ 1 $) that between the $ n $Covered coverings of $ X $ to the isomorphism of the covering of spaces and the conjugation classes of homomorphism between $ pi_1 (X) $ and $ S_n $ (from where $ S_n $ is the symmetric group), under the very bad assumption that $ X $ allows universal coverage. The proof of this fact (at least in Hatcher's book) is quite technical, if not transcendent.

If we focus on double linings, things simplify considerably, and we have a double cross between the double linings of $ X $ and $ Hom ( pi_1 (X), S_2) $; So after the universal set of coefficients we have a bijection

$$ {Double Coverings of X } longleftrightarrow H ^ 1 (X; mathbb {Z} / 2 mathbb {Z}). $$

Is there an easier way to prove this (natural) connection in this particular case?