Embedded surface in simply connected manifold is two sided? Suppose M is embedded surface(compact without boundary) in N^3 which is a simply connected three dimensional manifold, then why does M have to be two sided? Is there any example for this? I found a paper named "orientability of hypersurface in $R^n$", but I  don't quite understand the short argument. Thanks for you help. 
 A: This is a standard kind of argument that you can find in textbooks on differential topology. It works in any dimension, so $N$ can be an $n$-manifold and $M \subset N$ can be a compact submanifold of dimension $n-1$. I think probably Guilleman and Pollack have this argument. Here's a sketch.
Assume $M$ is one sided. It follows that $M$ has a one-sided component. By replacing $M$ with that component, we can assume that $M$ is connected.
Pick an arc $\alpha$ that hits $M$ transversely in a single point. Since $M$ is connected and one-sided, there exists an arc $\beta \subset N$ that is disjoint from $M$ and that stays in a regular neighborhood of $M$ and such that $\alpha \cap \beta = \partial \alpha = \partial \beta$. It follows that $\gamma = \alpha \cup \beta$ is a circle in $M$ that intersects $M$ transversely in exactly one point. By perturbing $\gamma$, we can assume it is smooth.
Suppose now that $N$ were simply connected. Then there would be a smooth function $f : D^2 \to N$ such that the restriction $f : S^1 = \partial D^2 \to N$ is a homeomorphism onto $\gamma$. By using general position arguments, after perturbing $f$ one may assume that in addition the map $f$ is transverse to $M$. It follows that $f^{-1}(M)$ is compact 1-manifold properly embedded in $D^2$, and so the set of endpoints of this 1-manifold forms an even number of points in $S^1$. From this it follows that the cardinality of the finite set $\gamma \cap M$ is even. But it's not, it's $1$.
