Yesterday, I believe, I proved a statement from algebraic topology without actually having properly read any materials about algebraic topology yet. So now, I just wanted to check if this is true, trivial or a known lemma, because I have been searching for it but didn't find anything. Here is the statement:
If $S$ is a smooth hypersurface on a sphere (a closed submanifold, without boundary, of codimension 1), then it is orientable. Further, if $S$ is connected, then its complement consists of two connected components, call them $X$ and $Y$. Further, if $S$ is simply-connected, then $X$ and $Y$ are both simply-connected too.
Some background
Initially, I was attempting to prove that the volume bounded by an immersion of $\mathbb{S}^2$ in $\mathbb{R}^3$ is homeomorphic to a 3-ball and I knew it would be a bit too hard, so for starters I tried proving that it is at least simply-connected. I realized that this still includes some dependency on homotopy groups (the fundamental group) and I know none of that topology. But then I realized that there is another statement which does not depend on simple-connectedness (of neither the surface nor the bounded volume) but is potent enough to carry the property of simple-connectedness down from the boundary to the volume. In fact, it works for any surface.
That another statement is this:
If $S$ is a smooth, compact hypersurface in $\mathbb{R}^n$ (a compact submanifold, without boundary, of codimension 1), then it is orientable. Further, if $S$ is connected, then its complement consists of two connected components, call them $X$ and $Y$; and exactly one of them is bounded, say $X$. For any $x\in X$, the space of all paths (a path is a continuous mapping from $[0,1]$ to $\mathbb{R}^n$) which begin at $x$, end at $S$ and are inside $X$ between the ends is connected (every two such paths are homotopy equivalent within the space of such paths).
[This is an answer to an earlier version of the title]. The claim is incorrect. For example, take the $n$-torus $T^n$ and drill out a small hole. You get a manifold with simply-connected boundary $S^{n-1}$ but the manifold itself is not simply-connected.
By van Kampen, since $S$ is simply-connected, the fundamental group of $S^n$ would be the free product of the fundamental groups of $X$ and $Y$. Since $\pi_1(S^n)$ is trivial for $n\geq2$, it follows the same for $X$ and $Y$.
https://en.wikipedia.org/wiki/Seifert%E2%80%93Van_Kampen_theorem