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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?

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    $\begingroup$ yes, this does look right. $\endgroup$
    – Thomas
    Commented Oct 11, 2015 at 8:11
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    $\begingroup$ Yes, and the same argument shows that $M^n$ is locally compact, locally contractible, etc. Just any local property of $\Bbb R^n$ ($\Bbb H^n$) is transferred to $X$ as $X$ is locally homeomorphic to $\mathbb R^n$ ($\Bbb H^n$). $\endgroup$ Commented Oct 11, 2015 at 12:10
  • $\begingroup$ This looks good. You can simplify it and avoid the use of excluded middle by just always taking the small open ball. Then you also get the stronger statement that it is locally contractible. $\endgroup$
    – saolof
    Commented Nov 23, 2022 at 11:57

2 Answers 2

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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.

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Now let’s focus on a smooth manifold $M;$ by definition we now that for any point $p \in M$ exists a neighborhood $U_{p}$ of $p$ and a homeomorphism $\phi\colon U_{p} \to \mathbb{R}^{d}.$

Since $\mathbb{R}^{d}$ (with the Euclidean topology) is locally path connected, $M$ inherits this property through $\phi,$ and we can conclude that a path connected manifold is connected and that if the manifold is locally path connected, then it is path connected if and only if it is connected.

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