No hypersurface with odd Euler characteristic Here is a classic problem which I encountered and could not solve:

Prove that a simply connected closed smooth manifold has no closed sub-manifold of co-dimension $1$ with odd Euler characteristic. 

Note: Closed means compact and without boundary. 
I would appreciate any comments or direction in solving this.
 A: 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.  
A: 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. 
