The intuitive idea is that the sphere connected the two manifolds is not contractible, which implies the (n-1)th homotopy group is not zero. Another argument, which I am not totally understand, uses the fact that the universal covering of an aspherical manifold has only one end. I am wondering if someone here could clarify these for me or give me a new argument.



How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical?
If $N$ is an open contractible manifold of dimension at least two, then it has one end. For instance, you can compute the zeroth reduced homology of $N$ minus a closed ball $B$ by using the long exact sequence for $(N,B)$ by using excision and the fact that $N$ is contractible.


If $M$ is a closed aspherical manifold of dimension at least three and $M$ is a connected sum of $M\_1$ and $M\_2$, then the fundamental group of $M$ is the free product of the the fundamental groups of $M\_1$ and $M\_2$. 


If $M\_1$ and $M\_2$ are closed aspherical manifolds of dimension at least three, then their fundamental groups are nontrivial, and so $\pi\_{1}(M)$ would be a nontrivial free product. But the universal universal cover of $M$ is one ended, and one ended groups do not decompose as nontrivial free products.


