# Open subsets of the connected sum $M_1\# M_2$ [duplicate]

I'm trying to solve a problem in John Lee's ITM (Problem 4-19), but seems that i need helps now. Here's the problem :

Let $M_1 \# M_2$ be a connected sum of $n$-manifolds $M_1$ and $M_2$. Show that there are open subsets $U_1,U_2 \subseteq M_1 \# M_2$ and points $p_i \in M_i$ such that $U_i \approx M_i \smallsetminus \{p_i\}$, $U_1 \cap U_2 \approx \mathbb{R}^n \smallsetminus \{0\}$, and $U_1\cup U_2 = M_1 \# M_2$.

The definition of the connected sum of $M_1\# M_2$ is the adjunction space of $M_1' \cup_f M_2'$ under a homeomorphism $f : \partial M_1' \to \partial M_2'$, where $M_i' = M_i \smallsetminus B_i$ and $B_i \subseteq M_i$ is a regular coordinate ball of $M_i$. Regular coordinate ball $B_i \subseteq M_i$ is a coordinate ball with homeomorphism $\varphi : B' \to \varphi(B')=B_{r'}(0)$, where $B' \supseteq B$, such that $\varphi(B) = B_r(0)$ and $\varphi(\bar{B}) = \bar{B}_r(0)$ for $r'>r>0$.

If we denote the embeddings as $e_i : M_i' \to M_1 \# M_2$, and the larger open subsets contain the coordinate balls $B_i$ as $B_i'$, i'm guessing that the desired open subsets are $U_1 = e_1(M_1')\cup e_2(B_2' \smallsetminus B_2)$ and $U_2 = e_1(B_1'\smallsetminus B_1) \cup e_2(M_2')$. But i having trouble to show that $U_i \approx M_i\smallsetminus \{p_i\}$. Actually i managed to show that (with $p_i$ is chosen as the "center" of coordinate ball $B_i$), but my solution is quite long. So i'm not sure it is correct (i'll add my solution here if needed).

If these choices of $U_1,U_2$ is correct i really appreciate if somebody tell me some hint to show that $U_i \approx M_i \smallsetminus \{p_i\}$. If it is not then i would like to know what the correct choices look like. Thank you.

$\textbf{Edit : }$I think i already got what i want here. It has a nice answer.

## marked as duplicate by Sou, Chris Custer, Joel Reyes Noche, The Phenotype, user223391 May 1 '18 at 20:59

In $M_i$ you have the open sets $B_i \subset B'_i \subset M_i$ as in your question. Choose $s < r$ and let $B_i'' = \varphi^{-1}(B_s(0))$. Define $V_i = M_i \backslash \overline{B_i''}$, $p_i = \varphi^{-1}(0)$ and $U_i = e_i(V_i)$. To see that $U_i \approx M_i \backslash \lbrace p_i \rbrace$ you construct a homeomorphism $h_i : V_i \to M_i \backslash \lbrace p_i \rbrace$ which is the identity on $M_i \backslash B_i$ and identifies $\overline{B}_i \backslash \overline{B_i''}$ with $\overline{B}_i \backslash \lbrace p_i \rbrace$ (the boundary must be fixed).
You can reduce this to showing that if $0 < s < r$ then $\overline{B}_r(0) \backslash \overline{B}_s(0) \approx \overline{B}_r(0) \backslash \lbrace 0 \rbrace$ with a homeomorphism fixing all points of the boundary of $\overline{B}_r(0)$. This can be done by radial stretching ($(s,r]$ stretches to $(0,r]$).