No hypersurface with odd Euler characteristic

A co-dimension 1 boundaryless, connected manifold in a simply connected manifold separates it into two components -- this is a version of the generalized Jordan-Brouwer separation theorem, and most proofs of this theorem adapt immediately to this case.

So your manifold, call it $N$ is the boundary of two manifolds $N = \partial W$, $N= \partial V$ and $V \cup W$ is the given simply connected manifold, and $V \cap W = N$, with both $V$ and $W$ path-connected.

So the next step is proving that if a manifold is the boundary of another manifold, its Euler characteristic is even. There's a lot of different arguments for this. A simple one is to start with $N = \partial V$, and construct the double $dV$ of $V$. Then $\chi dV = \chi V + \chi V - \chi N$.

So $$\chi N = 2 \chi V - \chi dV$$

Think about various cases. If $N$ is even dimensional, $dV$ is odd dimensional, and it's closed, so by Poincare duality, $\chi dV=0$ and you're done.

If $N$ is odd dimensional, $\chi N=0$ and you're done.

My solution after all:

Let $M$ be our manifold of dimension $m$ and suppose that $N \subset M$ be a hypersurface. Since $M$ is simply-connected, we obtain $H^1(M; \mathbb{Z})=0$ then by Poincare-duality, we also deduce that $H_{m-1}(M ;\mathbb{Z})=0$ which shows that $N$ is a boundary $N=\partial U$ where $U \subset M.$ Now, if we apply the Lefschetz duality to the pair $(U, \partial U)$, we obtain $\chi (N)=2 \chi (U)$ which proves the claim.