I recall that $M_1\#M_2$ is the connexe sum of two manifolds and it's defined as following: Let $B_1\subset M_1\backslash \partial M_1$ and $B_2\subset M_2\backslash \partial M_2$ where $M_i$ have dimension $n$ two set homeomorphic to $\bar{\mathbb B}^n=\{x\in\mathbb R^n\mid \|x\|\leq 1\}$. Let $f:\partial \bar{B_1}\longrightarrow \partial \bar{B_2}$ an homeomorphism. Then $$M_1\#M_2=(M_1\backslash B_1)\cup(M_2\backslash B_2)\Big/f.$$
I recall that $\mathbb S^n=\partial \mathbb B^{n+1}=\{x\in\mathbb R^{n+1}\mid \|x\|=1\}$.
Exercice : So, I have to show that if $M$ is a manifold of dimension $n$, then $$M\#\mathbb S^n=M.$$
My solution : Let $(U,\varphi)$ a chart of $\mathbb S^n$ and $(V,\psi)$ a chart of $M$ such that $\varphi(U)\supset \bar{\mathbb B}^n$ and $\psi(V)\supset \bar{\mathbb B}^n$. Let denote $B_1=\varphi^{-1}(\bar{\mathbb B}^n)$ and $B_2=\psi^{-1}(\bar{\mathbb B}^n)$. Let $$f=\psi^{-1}\circ \varphi|_{\partial B_1}$$ which is an homeomorphism $\partial B_1\longrightarrow \partial B_2$ (I think). Therefore $$M\# \mathbb S^n=(M\backslash B_2)\cup (\mathbb S^n\backslash B_1)\Big/f.$$
Question : How can I show that $\mathbb S^n\backslash B_1\cong B_2$ ?
This is the generalized Schönflies theorem.