# A manifold with boundary is locally (path) connected

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

$$\mathbb{H}^n= \{ (x_1,...,x_n) \in \mathbb{R}^n: x_n\geq 0 \}$$

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 $$U$$.
• Let me check if I got your answer right, because it's not entirelty clear to me why "$h(U \cap V)$ 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's my guess: we see $\mathbb{H}^n$ with the subspace topology inherited by $\mathbb{R}^n$, so an open set of $\mathbb{H}^n$ is an open set of $\mathbb{R}^n$ intersected with $\mathbb{H}^n$. – Leonardo Dec 3 '18 at 8:55
• Now, one can prove that the sets $W'$ of the form "open ball intersected with $\mathbb{H}^n$" form a basis for $\mathbb{H}^n$, so if $h(U \cap V)$ is open in $\mathbb{H}^n$ it's union of sets $W'$, in particular there exists a certain $W': h(x) \in W'$. I had to split the comment into two parts because I couldn't post it as a whole. – Leonardo Dec 3 '18 at 8:55