Classification of double coverings and $H^1(X ; \mathbb{Z}/2\mathbb{Z})$ It is a well known fact (see for example Hatcher's "Algebraic Topology", chapter $1$) that there is a bijection between the $n$-sheeted coverings of $X$ up to isomorphism of covering spaces and the conjugacy classes of homomorphisms between $\pi_1(X)$ and $S_n$ (where $S_n$ is the symmetric group), under the very poor assumption that $X$ admits universal covering. The proof of this fact (at least, the one presented in Hatcher's book) is quite technical,  although not transcendental. 
If we focus our attention on double coverings, things simplify quite a lot, and we have a bijection between the double coverings of $X$ and $Hom(\pi_1(X), S_2)$; by universal coefficient theorem, we have so a bijection
$$\{Double\ coverings\ of\ X\} \longleftrightarrow H^1(X; \mathbb{Z}/2\mathbb{Z}).$$
Does exist a simpler way to prove this (natural) bijection in this particular case?
 A: Using the universal coefficient theorem and the fact that $\mathbb{R}\mathrm{P}^\infty$ is a $K(\mathbb{Z}/2\mathbb{Z},1)$,
$$H^1(X;\mathbb{Z}/2\mathbb{Z})\cong \hom(H_1(X),\mathbb{Z}/2\mathbb{Z})\cong\hom(\pi_1(X),\mathbb{Z}/2\mathbb{Z})\cong[X,\mathbb{R}\mathrm{P}^\infty],$$
where the square brackets mean homotopy classes of maps $X\to \mathbb{R}\mathrm{P}^\infty$.
The space $\mathbb{R}\mathrm{P}^\infty$ has a two-fold cover that comes from the unit vectors in the so-called tautological line bundle: each point of projective space is a 1-D subspace of $\mathbb{R}^\infty$, so you can make a line bundle where the fiber over each point is the point as a vector space.  In particular, the quotient map $q:S^\infty\to \mathbb{R}\mathrm{P}^\infty$ is a two-fold cover.
Given a map $f:X\to \mathbb{R}\mathrm{P}^\infty$, you can pull back this two-fold cover to get one of $X$.$\require{AMScd}$
\begin{CD}
f^*(S^\infty) @>>> S^\infty \\
@VpVV @VVqV \\
X @>>f> \mathbb{R}\mathrm{P}^\infty
\end{CD}
This is independent of the choice of $f$ in its homotopy class.  Naturality is that maps $X\to Y$ give a contravariant map on covering spaces by pullback.
For the converse, we need additional assumptions on our spaces (or so I think: I don't have any counterexamples off the top of my head).  Let $p:X'\to X$ be a two-fold cover.


*

*If $X$ is paracompact (for example a manifold or a CW complex), then $p$ is the pullback of of $q$ for some map $X\to \mathbb{R}\mathrm{P}^1$.  (See Milnor-Stasheff, Theorem 5.6.  Promote the cover $p$ to a real line bundle $P:E\to X$ with $p^{-1}(x)$ being the two unit vectors in the fiber.)  This map then yields a first cohomology class by the above isomorphisms.

*If $X$ has a universal cover, then by the classification of covering spaces we know that each two-fold cover is in correspondence with a homomorphism $\pi_1(X)\to \mathbb{Z}/2\mathbb{Z}$.

If $X$ is a compact $n$-manifold, possibly with boundary, then by Poincare duality $H^1(X;\mathbb{Z}/2\mathbb{Z})$ is $H_{n-1}(X,\partial X;\mathbb{Z}/2\mathbb{Z})$.  At least when $n\leq 3$, such a homology class can be represented by a properly embedded codimension-1 hypersurface.  The two-fold cover comes from slicing the manifold along these hypersurfaces, then taking two copies of this and gluing them together along the newly exposed boundary.  If a loop in $X$ is orientation-reversing, then the loop must intersect the hypersurface an odd number of times, and a lift of the loop goes from one of the two copies to the other.
