Show that $M\#\mathbb S^n\approx M$. 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 $\mathbb S^n\backslash B_1\cong B_2$ ?
 A: This is the generalized Schönflies theorem.
A: Here's some details of the proof. Let $B_1 \subset M$ and $B_2 \subset \Bbb{S}^n$ be the chosen regular coordinate balls and denote $M' = M \smallsetminus B_1$ and $(\Bbb{S}^n)' = \Bbb{S}^n \smallsetminus B_2$ and  $f : \partial(\Bbb{S}^n)' \to \partial M'$ be some homeomorphism. Let $q : M' \sqcup (\Bbb{S}^n)' \to M \# \Bbb{S}^n$ be the quotient map. The connected sum $M \# \Bbb{S}^n$ is obtained from by cutting out an open ball $B_1$ and pasting back to the closed ball along the boundary sphere.

In short, to show that a connected sum of an any $n$-manifold $M \# \mathbb{S}^n$ is homeomorphic to $M$, we need to construct a quotient map $Q :  M' \sqcup (\Bbb{S}^n)' \to M$ that makes same identification as $q$. And then the result will follow from uniqueness of quotient topology.

Choose $B_2 \subseteq \Bbb{S}^n$ to be the open lower hemisphere (while $B_1$ arbitrary), so $(\Bbb{S}^n)'$ is the closed upper hemisphere, which is homeomorphic to a closed ball by stereographic projection. Denote $\sigma_N : \Bbb{S}^n \smallsetminus \{N\} \to \Bbb{R}^n$ and $\sigma_S : \Bbb{S}^n\smallsetminus \{S\} \to \Bbb{R}^n$ as the stereographic projection maps from the "north pole" $N =(0,\dots,0,1) \in \Bbb{R}^{n+1}$ and the "south pole" $S = (0,\dots,0,-1) \in \Bbb{R}^{n+1}$ respectively. 
Also by hypothesis $B_1 \subseteq M$ is a regular coordinate ball, so let $\psi : U \to B_R(0)$ be the homeomorphism and $U \supset \overline{B_1}$ be an open subset such that $\psi(U) = B_R(0) \supset \psi(\overline{B_1}) = \overline{\mathbb{B}^n}$ for some $R>0$ and $\psi(B_1) = \mathbb{B}^n$, $\psi(\partial B_1) = \Bbb{S}^{n-1}$.
First note that the open lower hemisphere $B_2\subseteq \Bbb{S}^n$ is a regular coordinate ball since we can choose an open subset $V \subseteq \Bbb{S}^n $ contain $\overline{B_2}$ defined by $$V :=\Bbb{S}^n \cap \{x \in \Bbb{R}^{n+1} \mid x^{n+1} < \varepsilon, 0 <\varepsilon <1\}$$ and a homeomorphism, obtained by restriction of $\sigma_N$ to $V$ such that $\sigma_{N}(V) = B_r(0) \supset \sigma_N(\overline{B_2}) = \overline{\mathbb{B}^n}$ and $\sigma_{N}(B_2) = \mathbb{B}^n$, $\sigma_{N}(\partial B_2) = \Bbb{S}^{n-1}$.
Let us define $\varphi_N := \sigma_N|_{\overline{B_2}}$. Also $(\Bbb{S}^n)' \approx \overline{\mathbb{B}^n}$ by $\varphi_S := \sigma_S|_{(\Bbb{S}^n)'} : (\Bbb{S}^n)' \to \overline{\mathbb{B}^n}$. Define $Q :  M' \sqcup (\Bbb{S}^n)' \to M$ by its restrictions to $M'$ and $(\Bbb{S}^n)'$ as $$Q|_{M'} = \iota_{M'} : M' \hookrightarrow M$$ and $$Q|_{(\Bbb{S}^n)'} := \iota_{\overline{B_1}} \circ \widetilde{f} \circ \varphi_N^{-1} \circ \varphi_S : (\Bbb{S}^n)' \to M.$$ 
The map $\iota_{\overline{B_1}}  : \overline{B_1} \hookrightarrow M $ is just inclusion map and $\widetilde{f} : \overline{B_2} \to \overline{B_1}$ is an extension of $f : \partial B_2 \to \partial B_1$ defined as $\widetilde{f} := (\psi|_{\overline{B_1}})^{-1} \circ g \circ \varphi_N$ and $g : \overline{\mathbb{B}^n} \to \overline{\mathbb{B}^n}$ defined as
    $$
 g(x)=
 |x| \hat{f} \big(\frac{x}{|x|}\big), \quad x\neq 0
    $$
and $g(0) = 0$. The map $\hat{f} : \Bbb{S}^{n-1} \to \Bbb{S}^{n-1}$ is the composition $$\hat{f} = \psi|_{\partial B_1} \circ f \circ (\sigma_N|_{\partial B_2})^{-1}.$$
The map $Q :  M' \sqcup (\Bbb{S}^n)' \to M$ continous surjective map, which is also a quotient map, since it's takes saturated open subsets to open subsets. It can be verify directly that $Q$ makes same identification as $q$. Therefore by uniqueness of quotient topology $M\#\Bbb{S}^n \approx M$.
