Let $M$ be a compact, connected, oriented $n$-dimensional manifold without boundary. Suppose that $M\#M\cong M$. Does it imply that $M \cong S^n$?
Sorry if this is a naive question. This is not my area, and I have very few examples of higher dimensional manifolds under my belt, so I wouldn't know how to construct a counter-example!
This is true. More generally, we prove that if $N\sharp M \cong N$ for a single closed orientable manifold $N$, then $M$ is homeomorphic to a sphere.
First, by the classification of compact surfaces, this is true when $N$ and $M$ are surfaces, so we may assume $n=\dim N \geq 3$.
Now, one can apply van Kampen's Theorem to learn that $\pi_1(M\sharp N)\cong \pi_1(M)\ast \pi_1(N)$. Since $N$ is compact, $\pi_1(N)$ is finitely generated, say with minimal generating set consisting of $r$ generators. Let $s$ be the minimal number of generators of $\pi_1(M)$. Then, $r+s$ is the size of the minimal generating set of $\pi_1(M)\ast \pi_1(N) \cong \pi_1(N)$, so $r+s = r$. Since $r$ is finite, this implies $s = 0$. Thus, $M$ is simply connected.
Using Mayer-Veitoris, and the fact that $N$ is orientable (assumption) and $M$ is orientable (since it's simply connected), one now sees that $H_i(N)\cong H_i(M\sharp N) \cong H_i(M)\oplus H_i(N)$ for $0< i < n$. Since the homology groups of $M\sharp N$ and $M$ are finitely generated abelian groups, we can cancel to find that $H_i(M) = 0$ for $0<i<n$.
All this shows that $M$ is a simply connected homology sphere. The Hurewicz theorem together with Whitehead's theorem now imply $M$ is homotopy equivalent to a sphere. Finally, the Poincare conjecture then implies that $M$ is homeomorphic to a sphere.
