Proof verification in Hatcher Algebraic Topology, Proposition 3.25 First, this is the link of the book, for convenience: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf#page=244
Proposition 3.25 If $M$ is connected, then $M$ is orietable iff $\tilde M$ has two components.
(Here $M$ is an $n$-manifold. The definition of $\tilde M$ is constructed above the statement.)
Proof. If $M$ is connected, $\tilde M$ has either one or two components since it is a two-sheeted covering space of $M$. If it has two components, they are each mapped homeomorphically to $M$ by the covering projection, so $M$ is orientable. Conversely, if $M$ is orientable, it has exactly two orientations since it is connected, and each of these orientations defines a component of $\tilde M$.
I think this proof is fulled of just claims. I took quite lot of time verifying each sentence. I almost understood all, but the only thing that I couldn't is in the last sentence that the two orientations of $M$ define the two components of $\tilde M$. That is, how do I have to show that $\tilde{M}$ is not connected? Thanks in advance.

Fix the two global orientations and call them $+1$ and $-1$. This gives you a map $\tilde M \to \{+1, -1\}$ which is continuous and surjective. So $\tilde M$ is disconnected. Conversely, the inverse images of $+1$ and $-1$ are each homeomorphic to $M$ so those are connected.
To see why the map is continuous you use the definition of the topology to show that $\tilde M$ is orientable (in particular it is locally orientable). So every point has a neighbourhood with a compatible orientation. Meaning $+1$ points have only $+1$ points in a neighbourhood of them and vice versa.
