If a smooth manifold X is covered by an odd sphere, then X is orientable.

Solution 1:

I assume you mean a smooth covering map. If so, recall that if $\tilde{X}$ is a connected, oriented, smooth manifold and $\pi:\tilde{X}\to X$ is a smooth normal covering map. Then $X$ is orientable if and only if every deck transformation is orientation preserving on $\tilde{X}$. (Lee, Smooth Manifolds, pg.392)

Say that $\pi:S^{2n+1}\to X$ is a smooth covering map. As $S^{2n+1}$ is simply connected, it is the universal cover of $X$ and is, in particular, a smooth normal covering map. Since $S^{2n+1}$ is connected and oriented, the theorem reduces to showing that every deck transformation is orientation preserving. Since the deck transformations act freely on $S^{2n+1}$, every non-identity deck transformation has no fixed points. Thus, in the spirit of Xipan's answer, every deck transformation is homotopic to the antipodal map, which has degree $1$. In particular, every deck transformation is orientation preserving.

Solution 2:

Any non-trivial deck transformation $f$ has no fixed point. $\forall p \in S^{2n+1}$, $p \ne f(p)$. Define $F: S^{2n+1}\times[0,1] \rightarrow S^{2n+1}$ by $$F(p,t)=\frac{(1-t)f(p)-tp}{|(1-t)f(p)-tp|}$$Since $p\ne f(p)$ this is well defined and smooth. Intuitively, since $p\ne f(p)$, there is a unique path( the shortest path, which is a geodesic) from $f(p)$ to the antipodal point $-p$, $F$ is just sliding $f(p)$ to $-p$ along this path. $F$ is a smooth homotopy so $\deg{f}=\deg(-id)=(-1)^{2n}=1$, so $f$ preserves orientation.

Push down this $f$ preserving orientation we get an orientation on the base space.