# connected manifolds are path connected

prove every connected manifold is path connected manifold .

my thought:

connected space : Let $X$ be a topological space. A separation of $X$ is a pair $U, V$ of disjoint nonempty open subsets of $X$ whose union is $X$. The space $X$ is said to be connected if there does not exist a separation of $X$ .

Components: Given $X$ , define an equivalence relation on $X$ by setting x~y if there is a connected subspace of $X$ containing both $x$ and $y$ . The equivalence classes are called the components (or the "connected components") of $X$.

Path Component: I define another equivalence relation on the space $X$ by defining x ~ у if there is a path in $X$ from $x$ to $y$ . The equivalence classes are called the path components of $X$ .

Theorem : The path components of $X$ are path connected disjoint subspaces of $X$ whose union is $X$ , such that each nonempty path connected subspace of $X$ intersects only one of them.

thank you so much

• There is a result that every connected and locally path-connected space is path connected. Can you show that the manifold is locally path-connected? Feb 12, 2015 at 17:48

Let $x\in X$. Consider $U:=\{y\in X\mathop{|}\text{there is a path from$x$to$y$}\}$. So $U$ is nonempty: $x\in U$. Claim: $U$ and $U^c:=X\smallsetminus U$ both are open. To prove this use the fact that given any point $z\in X$ there is a neighbourhood of $z$ which is homeomorphic to an open ball of $\mathbb{R}^n$ for some $n$. The homeomorphic image of a path connected set is path connected, so $U$ and $U^c$ both are open but $U \cup U^c = X$ which implies that $U=X$ since $U$ is nonempty.

• How exactly is concluded that $U$ and $U^c$ are open? I do not see it. Thanks. Jul 3, 2020 at 23:05

path connected sets are also connected, conversely is not always true (there is a famous counterexample which shows this fact). However a necessary condition To make the converse of the previous statement possible is that the connected sets are required to be locally Euclidean. So a topological manifold, by definition is also a locally Euclidean space, if you can prove the previous statement you can get the answer. Hope it helps

Given our manifold M is connected, let $$x, y \in M$$ be arbitrary point.

Since for any $$x$$ there is an open set $$U_x \cong \mathbb{R}^n$$ for which $$x$$ is contained, let $$C_x$$ denote the path component of $$x$$. If $$y$$ is contained then we are done. However, if not then as then let $$\bigcup_{x \in C_x}U_x =C_x$$ which is open.

$$M$$ is the union of path connected components, by previous reasoning all path connected components are open. Thus this implies that $$M$$ is disconnected contradicting our hypothesis $$y \notin M$$. Therefore, $$M$$ is path connected.

Conversely, path connectednesss implies connectedness as required.

Proving that Manifold is connected if and only it is path connected.