To show $\overline{B^n}$ is a $n$-manifold with boundary, apparently there is a trick to use stereographic projection after subtracting out the radius connecting $0$ to the north pole.
I'm familiar with the geometric interpretation of stereographic projection of $S^n-N$, but not with a closed ball in $\mathbb{R}^n$. What is the stereographic projection for a closed ball?
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You're probably looking at the first edition of Introduction to Topological Manifolds. I realized that my hint to "use stereographic projection" was not very helpful, so in the second edition I expanded on it:
Consider the map $\pi\circ\sigma^{-1}\colon\mathbb R^n\to \mathbb R^n$, where $\sigma$ is the stereographic projection and $\pi$ is a projection from $\mathbb R^{n+1}$ to $\mathbb R^n$ that omits some coordinate other than the last.
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Let $\sigma$ and $\sigma^{-1}$ be given as in p.25, l.4 and l.6 in https://wj32.org/wp/wp-content/uploads/2012/12/Introduction-to-Smooth-Manifolds.pdf. [The simple geometric ideas behind their construction] I. Let $f_1$ be the stereographic projection of $S^{n-1}$ onto the hyperplane $u_n=0$. Then $f_1(X_1,\cdots,X_n)=(\frac{X_1}{1-X_n},\cdots,\frac{X_{n-1}}{1-X_n},0)=(u_1,\cdots,u_{n-1},0)$ and $f_1^{-1}(u_1,\cdots,u_{n-1},0)=(\frac{2u_1}{\|\tilde{u}\|^2+1},\cdots, \frac{2u_{n-1}}{\|\tilde{u}\|^2+1},\frac{\|\tilde{u}\|^2-1}{\|\tilde{u}\|^2+1})$, where $\tilde{u}=(u_1,\cdots,u_{n-1})$ [p.9, 6.2 & 6.4 in J.B. Conway's Functions of One Complex Variable, 2nd ed.].\ II. Let $f_r (r> 0)$ be the stereographic projection of $\partial \bar{B}(0,r)$ onto the hyperplane $u_n=0$. Then $f_r(r(X_1,\cdots,X_n))=r(\frac{X_1}{1-X_n},\cdots,\frac{X_{n-1}}{1-X_n},0)=r(u_1,\cdots,u_{n-1},0)$ and $f_r^{-1}(r(u_1,\cdots,u_{n-1},0))=r(\frac{2u_1}{\|\tilde{u}\|^2+1},\cdots, \frac{2u_{n-1}}{\|\tilde{u}\|^2+1},\frac{\|\tilde{u}\|^2-1}{\|\tilde{u}\|^2+1})$.\ III. Let $(x_1,\cdots,x_n)\in \bar{B}(0,1)\setminus (\{0\}^{n-1}\times [0,1])$. Then $(x_1,\cdots,x_n)=\|x\|(X_1,\cdots,X_n)$.\ $\sigma (\|x\|(X_1,\cdots,X_n))=(\|x\|)^{-1}f_{\|x\|}(\|x\|(X_1,\cdots,X_n))+(0,\cdots,0, \|x\|^{-1}-1)\\ =f_1(X_1,\cdots,X_n)+(0,\cdots,0, \|x\|^{-1}-1)=(u_1,\cdots,u_{n-1},u_n)$, where $u_n=\|x\|^{-1}-1$.\ $\sigma^{-1}(u_1,\cdots ,u_n)=\|x\|f_1^{-1}(u_1,\cdots,u_{n-1},0)=\frac{1}{u_n+1}f_1^{-1}(u_1,\cdots,u_{n-1},0)$.
Remark 1. $u_n$ in III is used as a translation and a label for the reminder that reminds us of the size of the ball for the stereographic projection. The reminder greatly helps us find $\sigma^{-1}$.\ Remark 2. For the philosophy of this passage, read §28 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-proofs
Think of the ball as being foliated by concentric spheres of different radii. Removing the radius connecting the center to the north pole removes the north pole from each of them. Now project each of them stereographically onto $\mathbb{R}^{n-1}\times\{x_n\}\subset\mathbb{R}^{n}$ in such a way that the boundary of the ball corresponds to $x_n=0$ and $x_n$ grows to $\infty$ as the radius approaches $0$. This gives you a chart that maps the ball with a radius removed onto the upper half-space with the boundary sphere being mapped into its bounding hyperplane $\mathbb{R}^{n-1}\times\{0\}$.