Closure of regular coordinate ball is homeomorphic to the closure of its Euclidean ball?

46 Views Asked by At

Lee in Introduction to Topological Manifolds defines regular coordinate balls as follows:

"Let $M$ be a $n$-manifold. We say that a coordinate ball $B\subset M$ is a regular coordinate ball if there is a neighbourhood $B'$ of $\overline{B}$ and a homeomorphism $\varphi:B'\to B_{r'}(x)\subset \mathbb{R}^n$ that takes $B$ to $B_r(x)$ and $\overline{B}$ to $\overline{B_r(x)}$ for some $r'>r>0$ and $x\in\mathbb{R}^n$."

I was wondering if this means that $B\cong B_r(x)$ and $\overline{B}\cong\overline{B_r(x)}$? I know that restricting $\varphi$ to $B$, I get a continuous injective map $B\to B_{r'}(x)$, which gives a continuous bijection $B\to B_r(x)$. Here I used $f:X\to Y$ continuous, $A\subset X$ implies $f|_{A}:A\to f(A)$ is continuous. Similarly, I can show there is a continuous bijection $\overline{B}\to \overline{B_r(x)}$ upon restricting $\varphi$ to $\overline{B}$.

I am having trouble showing that the inverse map is also continuous. Any suggestions will be greatly appreciated!!