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

  • 6
    $\begingroup$ 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? $\endgroup$ Feb 12, 2015 at 17:48

3 Answers 3


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.

  • 1
    $\begingroup$ How exactly is concluded that $U$ and $U^c$ are open? I do not see it. Thanks. $\endgroup$
    – Cornman
    Jul 3, 2020 at 23:05
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    $\begingroup$ @Cornman, if every $z\in X$ belongs to some path-connected open, then every $z\in U$ belongs to a path-connected open fully included in $U$ (otherwise it would connect some $y\in U^c$ to $z$ through a path, contradiction), and for the same reason every $y\in U^c$ belongs to a (path-connected) open fully contained in $U^c$. Thus $U$ and $U^c$ are open by definition. $\endgroup$
    – Albert
    Mar 9, 2023 at 13:26

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.


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