# For path connected manifolds, the orientation double cover is path connected iff manifold is not orientable

The "orientation double cover" $\tilde{M}$ of a n-manifold $M$ is defined as follows:
$\tilde{M} = \{\mu_x | x \in M, \mu_x \textrm{ is a generator of } H_n(M,M-\{x\})\}$.
Well, if we assume that the n-manifold $M$ is path connected, apparently it is true that $\tilde{M}$ is path connected if an only if $M$ is not orientable. I've been given an explanation for this too, albeit terse: "$\tilde{M}$ is path connected iff $\tilde{M} \to M$ is a non-trivial 2-sheeted covering map iff $\tilde{M} \to M$ has no sections iff $M$ is non-orientable".
Save for the following implication: IF $\tilde{M}$ is path connected THEN $\tilde{M} \to M$ is a non-trivial 2-sheeted covering map, I cannot see where any of the other implications come from! It all feels quite unintuitive. Why is a section the same thing as an orientation? Why would a non-trivial double cover mean that the 2:1 surjection $\tilde{M} \to M$ has no sections? Perhaps by understanding some of these implications I would be able to understand the rest...but I'm struggling to see why any of this is true.

Most of the question has nothing to do with orientations. Namely, if $p\colon Y\to X$ is a double-cover of a sufficiently nice, connected topological space $X$, then the following are equivalent:

1. The cover $p\colon Y\to X$ is trivial, i.e., as a cover isomorphic to the trivial one $\tau\colon X\amalg X\to X$.
2. The map $p\colon Y\to X$ admits a continuous section.
3. The space $Y$ is not connected.

There is a dry general topology proof of this playing around with closed and open sets and so on, or we could throw the theory of principal bundles at it, but it seems the OP wants some intuition, so I will try something more basic.

First of all, the trivial double cover $\tau\colon X\amalg X\to X$ has exactly two distinct continuous sections, namely, the two natural inclusions $X\to X\amalg X$. Thus, if $p$ is equivalent to $\tau$, then $p$ also has (two distinct) global continuous sections. This shows that the first statement implies the second.

Now we show that the second statement implies the third. Since $p$ is a double cover, we may fix, for each $x\in X$, an open neighbourhood $U_x\subset X$ of $x$ such that $p^{-1}U_x = V_x\amalg V_x'$ for some open subsets $V_x,V_x'\subset Y$ such that $p$ restricts to isomorphisms $p_x = p|_{V_x}\colon V_x\to U_x$, $p'_x = p|_{V'_x}\colon V_x'\to U_x$. Notice that $p|_{p^{-1}U_x}\colon V_x\amalg V_x'\to U_x$ has exactly(!) two continuous sections, namely, the inverses of $p_x$ and $p'_x$. In particular, if there exists a global continuous section $s\colon X\to Y$, then its restriction to $U_x$ equals either $(p_x)^{-1}$ or $(p'_x)^{-1}$ and so either $s(U_x) = V_x$ or $s(U_x) = V'_x$, depending on whether $s(x)\in V_x$ or $s(x)\in V_x'$. Reordering if necessary, we may assume that $s(U_x) = V_x$ for every $x\in X$ and we then get $s(X) = \bigcup_{x\in X}s(U_x) = \bigcup_{x\in X}V_x$ and $Y-s(X) = \bigcup_{x\in X}V'_x$; both of these sets being open and non-empty, this shows that $Y = s(X)\amalg(Y-s(X))$ has two components.

Finally, assuming $Y = Y_1\amalg Y_2$ with $Y_1,Y_2\not=\emptyset$, we see that each each $p|_{Y_i}\colon Y_i\to X$, $i=1,2$, is a one-sheeted cover, hence an isomorphism and so $p$ is trivial. That is, the third statement implies the first.

For the rest of the question, I don't really know what to say since for me, an orientation is exactly the choice of a global section: for every $x\in M$ a generator $H_n(M,M-x)$ which varies continuously with $x\in M$.