For a simply connected $n$-manifold $M\subseteq\Bbb{R}^k$, I want to show that $M$ is orientable.
Take a point $p\in M$ and take an $n$-disc, $D^n$, around $p$ (we can take it as small as we please). Since $S^{n-1}$ is orientable and $M$ (and consequently $TM$) is simply connected, the orientation map $S^{n-1}\to TM$ can be extended to $D^n\to TM$. So around every point there is such a disc. We can construct an atlas (an orientation) out of these.
Does this suffice to prove the claim?
Best Answer
Let $M$ be a connected non-orientable manifold of dimension $n$ and let $M^*$ be its oriented double cover which is connected as $M$ is non-orientable. By general covering space theory, there is a short exact sequence $$0\to\pi_1(M^*)\to \pi_1(M)\stackrel{f}{\to}\mathbb{Z}/2\mathbb{Z}\stackrel{g}{\to} 0$$ and so if $\pi_1(M)$ is trivial, then $\mathbb{Z}/2\mathbb{Z} = \ker g =\operatorname{im}f = 0$ which is a contradiction.