18
$\begingroup$

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

$\endgroup$
1
  • 4
    $\begingroup$ 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? $\endgroup$ – Stefan Hamcke Feb 12 '15 at 17:48
25
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ How exactly is concluded that $U$ and $U^c$ are open? I do not see it. Thanks. $\endgroup$ – Cornman Jul 3 '20 at 23:05
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.