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Proposition 1.11. Let M be a topological manifold. (a) M is locally path-connected. (b) M is connected if and only if it is path-connected. (c) The components of M are the same as its path components. (d) M has countably many components, each of which is an open subset of M anda connected topological manifold.

Proof. Since each coordinate ball is path-connected, (a) follows from the fact that M has a basis of coordinate balls. Parts (b) and (c) are immediate consequences of (a) and Proposition A.43. To prove (d), note that each component is open in M by Proposition A.43, so the collection of components is an open cover of M . Because M is second-countable, this cover must have a countable subcover. But since the components are all disjoint, the cover must have been countable to begin with, which is to say that M has only countably many components. Because the components are open, they are connected topological manifolds in the subspace topology

I have a doubt about the last sentence. For a component to be a manifold I need it to be Hausdorf, second countable and locally euclidean. The first two are properties of the subspace topology. For the third one, How do we use the fact that a component is open in order to prove that it is locally euclidean?

some_math_guy
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2 Answers2

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A connected subset of a manifold need not be a topological manifold. For instance, the top left corner point of the Warsaw circle does not have any connected neighbourhood (in the subspace topology).

The Warsaw circle

It follows that the Warsaw circle isn't locally euclidean, although it is a (closed, even compact) subspace of $\Bbb R^2$.

Another counter example: the union of the vertical and the horizontal axes of $\Bbb R^2$ is a closed subset which is not a topological manifold: no neighbourhood of the origin is homeomorphic to any euclidean open subset.

Of course, being open is not a necessary condition. The horizontal line in $\Bbb R^2$ is locally euclidean (it is homeomorphic to $\Bbb R$) although it is not open in $\Bbb R^2$.

It turns out (see the comment section) that OP is confused about this fact: an open subset of a locally euclidean space is locally euclidean. Here is a short proof. Let $U\subset M$ be an open subset, with $M$ locally euclidean. Let $x\in U$. Since $M$ is locally euclidean, there exist $V\subset M$ an open neighbourhood of $x$ in $M$, and a map $\varphi\colon V\to \Bbb R^n$, which is an homeomorphism onto its image. Then $V\cap U$ is an open neighbourhood of $x$ in $U$ which is homeomorphic to $\varphi(U\cap V)$, which is itself an open subset of $\Bbb R^n$ (since $\varphi$ is an homeomorphism onto its image). It follows that $U$ is locally euclidean.

Didier
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  • Nice counterexamples. What about the following reasoning as a direct proof: If I take a point $x$ of a component C of the Manifold M, then if C is open, by definition of openess there is a open neighborhood $U_x$ contained in C, since open sets in the subspace topology are of the form $C \cap V$, with $V$ an open set of $M$ , $\exists V $ open set of $M$ such that $U_x = C \cap V$. For$ C \cap V$ to be homeomorphic to an open subset of R^n, C must be open, so that the intersection of open sets C and V is open. Otherwise there is no guarantee that $C \cap V$ is open. Would this work? – some_math_guy Oct 25 '22 at 13:21
  • @mathlover I don't really get your point. I was just answering your question "Can a subset of a manifold be non locally euclidean" by a counterexample. But there surely are non-open subsets that are locally euclidean, and we call them submanifolds – Didier Oct 25 '22 at 13:34
  • I was trying to prove that a component is locally euclidean using the fact that the component is an open set in the topology of the whole manifold – some_math_guy Oct 25 '22 at 14:06
  • I think I did not state my question clearly in the title. My question was actually "why do we need a component to be open to be a manifold IN ORDER TO PROVE THAT THE COMPONENT IS LOCALLY EUCLIDEAN". I should have stated it like that – some_math_guy Oct 25 '22 at 14:08
  • Got it. This is not needed as in "if and only if". Openness is a sufficient condition. – Didier Oct 25 '22 at 14:10
  • Exactly, I want to prove openness $\implies$ locally euclidean. Is my argument in the first comment correctly proving this or how would I do that? – some_math_guy Oct 25 '22 at 14:12
  • @mathlover See my edit – Didier Oct 25 '22 at 14:15
  • OK I see I needed to use the fact that the restriction of a homeomorphism to an open subset is still a homeomorphism. And the fact that the component U is open was importat to state that $U \cap V$ is open, right? – some_math_guy Oct 25 '22 at 14:37
  • It is important to state that $\varphi(U\cap V)$ is open in $\Bbb R^n$. – Didier Oct 25 '22 at 14:40
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Components of general topological spaces need not be open (they are always closed, but this is irrelevant here). However, in locally connected spaces (and hence in particular in locally path connected spaces) the components are known to be open. This is proved in Lee's book. See also About locally path-connected spaces.

Kritiker der Elche
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  • Yes, I saw that proof, my question is if that had not been the case, why is the statement "Because the components are open, they are connected topological manifolds in the subspace topology" false? I meant why do we need openess of the components to conclude it is locally euclidean, can't I just conclude that without using the fact that the component is open?and only rely in for instance that if at the whole manifolds is locally eucilidean, so must be it for each component? – some_math_guy Oct 25 '22 at 12:40
  • @mathlover The question "Why wouldn't it hold if the component was not an open set?" does not make sense since components are open. Perhaps you wanted to ask why an arbitrary connected subset of a manilold $M$ needs not be locally Euclidean? If so, you should edit your question. – Kritiker der Elche Oct 25 '22 at 15:21
  • I have edited my question to reflect my actual question – some_math_guy Oct 25 '22 at 17:56