I'm trying to prove that a topological manifold with or without boundaries is locally (path) connected.
I think I've done the manifold without boundary part: a manifold without boundary is locally Euclidean, so it admits a basis of coordinate balls, which are homeomorphic to open balls, which are path connected because convex. Because path connectedness implies connectedness and it is preserved by continuous functions it follow that a manifold admits a basis of (path) connected sets.
I can't come up with nothing useful for the "with boundary" part. My intuition was to somehow apply the same line of reasoning as above, knowing that for a manifold with boundary every point has a neighbourhood that is either homeomorphic to $\text{int} \mathbb{H}^n$ or to an open subset of $\mathbb{H}^n$ including $ \partial\mathbb{H}^n$. My naive idea would be to show that for the first case one can use again the fact that an open ball is (path) connected, and an open ball which intersects $ \partial\mathbb{H}^n$ is still convex, thus (path) connected.
Can somebody tell me whether I should proceed on this line of reasoning or if I should come up with something different?
Thanks in advance
$\mathbb{H}^n= \{ (x_1,…,x_n) \in \mathbb{R}^n: x_n\geq 0 \}$
Best Answer
Your arguments are correct. Let $M$ be manifold (with or without boundary), $x \in M$ and $U$ be an open neighborhood $x$. There exists an open neigborhood $V$ of $x$ and a homeomorphism $ h : V \to W$, where $W$ is an open subset of $\mathbb{H}^n$. $h(U \cap V)$ is an open neighborhood of $h(x)$ in $\mathbb{H}^n$, hence there exists $\epsilon > 0$ such that $B_\epsilon(h(x)) \cap \mathbb{H}^n \subset h(U \cap V)$. Here $B_\epsilon(y_0)= \{ y \in \mathbb{R}^n \mid \lVert y - y_0 \rVert < \epsilon \}$. But $W' = B_\epsilon(h(x)) \cap \mathbb{H}^n$ is the intersection of convex sets, hence itself convex and therefore path connected. We conclude that $h^{-1}(W')$ is an open path connected neighborhood of $x$ which is contained in $U$.