I'm trying to prove that every topological manifold $M^n$ (with or without boundary) is locally path-connected. My attempt:
(Without boundary): Let $x\in M^n$ and $V$ be any open set containing $x$. Let $(U,\phi)$ be a chart containing $x$, so $U\cap V$ is also open, and $U\cap V$ is homeomorphic to $\phi(U\cap V)\subset\mathbb{R}^n$, which may or may not be path-connected. If it is path-connected, then $U\cap V= \phi^{-1}(\phi(U\cap V)\subset V$ is path connected, since path-connectectedness is preserved under homeomorphism and we are done. If $\phi(U\cap V)$ is not path-connected, then since it is open, we can find a sufficiently small $r$ so $B_r(\phi(x))\subset \phi(U\cap V)$. $B_r(\phi(x))$ is path-connected, so $x\in\phi^{-1}(B_r(\phi(x)))\subset V$ is path-connected.
(With Boundary): Let $\mathbb{H}^n=\{(x^1,…,x^n)\in\mathbb{R}^n:x^n=0\}.$ Replace $\mathbb{R}^n$ with $\mathbb{H}^n$ and $B_r(\phi(x))$ with $\mathbb{H}^n\cap B_r(\phi(x))$ in the argument above.
Does this look right?
Best Answer
Another way and by the definition of locally path connected, suppose $p$ a point from $M$ manifold, and $U_{p}$ is ball and $(f,U_{p})$ is a chart. So since $\mathbb{R}^{n}$ is locally path connected there exist $V_{p}$ neighborhood such that $V_{p}$ is in $U_{p}$ and $V_{p}$ is path connected.