Definition of a manifold

Question

NOT A DUPLICATE: Homeomorphism in the definition of a manifold for example is slightly different.

A manifold according to Wikipedia, a book by Spivak, and serveral other books has the following in common:

$\forall x\in M\exists n\in\mathbb{N}_{\ge 0}\exists U$ neighbourhood of $x$ such that $U$ is homeomorphic to (an open subset) of $\mathbb{R}^n$

I don't like the "neighbourhood" part as:

A set $U\subseteq M$ is a neighbourhood to $x$ if $\exists V$ open in $M$ with $[x\in V\wedge V\subseteq U]$

There is no requirement for $U$ to be open. It could be closed!

Problem:

Suppose that $U$ is homeomorphic to $\mathbb{R}^n$ by a function $f:U\rightarrow\mathbb{R}^n$, we know that $f$ is bijective and continuous by definition. This means it is surjective. Thus $f^{-1}(\mathbb{R}^n)=U$

By continuity of $f$ this means that $U$ is open in $M$

This is a contradiction, as $U$ need not be open.

I would be much happier if the definition was "there exists an open set containing $x$ that is homeomorphic to $\mathbb{R}^n$

The open subset of $\mathbb{R}^n$ part

The definition requires there exists a neighbourhood (not all neighbourhoods) homeomorphic to $\mathbb{R}^n$ is this the same as requiring the neighbourhood be homeomorphic to an open subset of $\mathbb{R}^n$?

Thoughts:

I understand that any open interval (in $\mathbb{R}$) is homeomorphic to all of $\mathbb{R}$ however the union of two distinct intervals is open but not homeomorphic to all of $\mathbb{R}$, using this sort of logic it suggests that:

I require (probably through the Hausdorff property) the ability to find a small enough connected (I suspect) open set. Then the two would be equivalent.

I could prove this if I assume the manifold has a countable topological basis (because then it is metricisable and I can use open balls) but I'd like to prove it for all manifolds.

You are correct that neighborhood is used inconsistently amongst multiple sources. Wikipedia, being written by the masses, is going to have inconsistencies like this. However, I will guess that a well crafted textbook, such as Spivak's, will use the word "neighborhood" in a way that avoids this inconsistency. — Lee Mosher, Sep 13 '15 at 19:47
@LeeMosher to my knowledge it doesn't define it. Every time I've encountered "neighbourhood" it usually means "any subset" and they even use neighbourhood systems as a precursor to topologies. (Spivak's book on differential geometry assumes a lot of knowledge, which I do have! I just want to resolve the ambiguities now) - please don't forget there is a second part to the question — Alec Teal, Sep 13 '15 at 19:48
Look at the solution to exercise 57 here: http://math.fau.edu/kalies/mtg6316/exercise_solns51-100.pdf . It basically shows how just about everything about the two definitions of neighborhoods agree — Alan, Sep 13 '15 at 19:57
@Alan that's a great link actually. I'm going to try and use that now to show that the definitions actually don't matter. Having said that I'm not having the easiest time and would love your take an on an answer (you seem to get I want to do it formally) — Alec Teal, Sep 13 '15 at 21:12
@AlecTeal Those were from when I took general topology last fall, we each worked on separate problems then the solutions got put together and shared for the class. One of the best classes I've had. I'll take a look later when I have more time/focus. — Alan, Sep 13 '15 at 21:26

Rob Arthan · Answer 1 · 2015-09-13T22:21:30.730

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The problem in the box you've labelled "Problem" is not a problem: the chart $f$ is required to be continuous on $U$, but not on $M$, so $f^{-1}(\mathbb{R}^n)$ could be closed in $M$ (although it will always be open too). The topological subspace $\{0, 1\} \times (0, 1)$ of $\mathbb{R}^2$ is a $1$-manifold in which every point has a neighbourhood that is homeomorphic with $\mathbb{R}$ and is both open and closed.

It doesn't make any essential difference in the definition of an $n$-manifold whether you require the domains of the charts to be neighbourhoods or open neighbourhoods or whether you require the ranges of the charts to be open subsets of $\mathbb{R}^n$ or the whole of $\mathbb{R}^n$: any point in an open subset of $\mathbb{R}^n$ has a neigbourhood that is homeomorphic with the whole of $\mathbb{R}^n$; this means that if $M$ is a topological space and $x$ is a point of $M$, then $x$ has a neighbourhood homeomorphic with $\mathbb{R}^n$ iff for some open $V \subseteq \mathbb{R}^n$, $x$ has a neighbourhood homeomorphic with $V$.

edited Sep 13 '15 at 22:21

answered Sep 13 '15 at 20:11

Rob Arthan

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The last bit "neighbourhood that is homeomorphic to $\mathbb{R}^n$" is only half the proof. I am fine with that direction. I want to be convinced that these definitions are the same. – Alec Teal Sep 13 '15 at 20:17
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Have you got a proof that the preimage of a chart will be both open and closed in the manifold? http://math.stackexchange.com/questions/872659/on-different-definitions-of-neighbourhood this is more useful I think, which I believe shows (not read it all yet) that the definitions of neighbourhood are equivalent. – Alec Teal Sep 13 '15 at 20:18
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Last comment, we're not talking about an $n$-manifold. I really would love a proof of what you've said (and the second half of what you address in the second paragraph) – Alec Teal Sep 13 '15 at 20:20
For your second comment: the domain of a chart need not always be open and closed, but it will always be open (if your definition of manifold requires the range of the chart to be open in $\mathbb{R}^n$). – Rob Arthan Sep 13 '15 at 20:25
Not the downvoter BTW. (I'd rather this be productive) I know already that these manifolds are not different because authors from both sides of the neighbourhood divide for example use it. There's a comment above and I've commented here showing something that seems to unite the two. I'm looking to be formal about this. The nearest you've come is half a proof in the second paragraph. I also do want the bit I mention above where I think the Hausdorff property plays a part. We must know there's a smaller open set somehow to complete the missing half of your proof (easy in T2 manifolds) – Alec Teal Sep 13 '15 at 20:32
Please can you state what this "missing half" is? – Rob Arthan Sep 13 '15 at 20:33
Sure, so we have two definitions. One states that things (neighbourhoods in some form) must be homeomorphic to $\mathbb{R}^n$ the other states that it must be homeomorphic to an open subset of $\mathbb{R}^n$ - it is obvious that an open subset of $\mathbb{R}^n$ may not be homeomorphic to an open subset (proof by contradiction: $\mathbb{R}$ and $(0,1)\cup(2,3)$ are not homeomorphic) HOWEVER if $(0,1)\cup(2,3)$ is our neighbourhood of $x$ then somehow we can "separate it" to just one of the intervals. -> – Alec Teal Sep 13 '15 at 20:36
You've touched on this, but I wouldn't call it a proof. Also the other direction is trivial. As $\mathbb{R}^n$ is an open subset of itself. – Alec Teal Sep 13 '15 at 20:38
I've expanded the answer. The point is that you you just have to focus at the point of interest. – Rob Arthan Sep 13 '15 at 20:41
I've read the edit and this is where I doubt you. If we had a countable basis to the topology (which we do not, although some authors define manifolds as having a countable basis) then I would be inclined to agree, because we can 'get that smaller' neighbourhood that is homeomorphic to $\mathbb{R}^n$ (every T2 manifold is also metrisisable so a countable basis is just the balls of radius 1/n say - not sure on the exact proof, but I can believe the claim) we only have the Hausdorff property to use. (read both edits before posting) – Alec Teal Sep 13 '15 at 20:43
Let $f$ be the homeomorphism of a neighbourhood of $x$ in $M$ with an open subset $V$ in $\mathbb{R}^m$. Pick a neighbourhood $W$ of $f(x)$ in $V$ that is homeomorphic with $\mathbb{R}^n$, then $f^{-1}(W)$ is the neighbourhood of $x$ you are looking for in $M$. – Rob Arthan Sep 13 '15 at 20:48
Okay good! But the proof that such a $W$ exists.... – Alec Teal Sep 13 '15 at 20:52
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@AlecTeal: Open balls in Euclidean space are homeomorphic to $\Bbb R^n$. Such a $W$ exists by the definition of $V$ being open. – Sep 13 '15 at 20:54
My goal here is to show concretely that the definitions of manifold are all equivalent. I cannot stress enough how not helpful just saying it is. Like take the $(0,1)\cup(2,3)$ example - how do we make this into just $(0,1)$ or $(2,3)$ using the hausdorff property. I can see that one does exist but I cannot prove it! – Alec Teal Sep 13 '15 at 20:54
@MikeMiller of course! Wish I'd seen that sooner. Thanks! (this shows one exists, I now have to carefully show either definition of neighbourhood exists, and then show a homeomorphism, because $(0,1)\cup(2,3)$ isn't homeomorphic to (open ball in it) - but I can find one that exists) – Alec Teal Sep 13 '15 at 20:54
@RobArthan I'm looking into the claim "a neighbourhood (as any set containing an open set) with the subspace topology" being the same as "an open set" when it comes to being homeomorphic to (something to do with $\mathbb{R}^n$) - you dismiss this as not being a problem.... – Alec Teal Sep 13 '15 at 21:01
I really don't want to open a second question. I don't see how $\exists$ neighbourhood (in the not open but contains an open set sense) which is homeomorphic to $\mathbb{R}^n$ $\implies$ there's an open set which is homeomorphic to $\mathbb{R}^n$ - I need a bigger set for it to be a subspace of! – Alec Teal Sep 13 '15 at 21:07
The point $x$ in $M$ is in the interior of this neighbourhood $U$ of $x$ that is the domain of the chart $f$ (that's what it means for $U$ to be a neighbourhood of $x$). The chart maps $U$ homeomorphically to an open subset $V$ of $\mathbb{R}^n$ and $V$ contains an open neighbourhood of $f(x)$ that is homeomorphic to all of $\mathbb{R}^n$. – Rob Arthan Sep 13 '15 at 21:42
@RobArthan I've done it. Half of it on paper, the other half I can see but haven't done. See my answer if you're curious as to what I wanted. (I've not filled it in yet, but I've stated what I now know to be true.) – Alec Teal Sep 13 '15 at 22:29
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I understood your question when I first read it. If you are happy that you can now answer it, good for you. – Rob Arthan Sep 13 '15 at 22:46

Alec Teal · Accepted Answer · 2015-09-13T22:26:55.657

For a topological manifold, $M$ with topology $\mathcal{J}$

I've done half of the question. I have shown that:

$\forall x\in M\exists V\text{ neighbourhood of }x\exists n\in\mathbb{N}_{\ge0}[x\in V\wedge V\cong\mathbb{R}^n]$
$$\iff$$ $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

Which is the first half done.

I will write the proof in later, I'm doing other things right now, but this is proved, no handwaving - formally proved!

The second part is that:

$\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}\exists V[V\in\mathcal{O}(\mathbb{R}^n)\wedge x\in U\wedge U\cong V]$ $$\iff$$ $\forall x\in M\exists U\in\mathcal{J}\exists n\in\mathbb{N}_{\ge 0}[x\in U\wedge U\cong\mathbb{R}^n]$

I've outlined this but not completed it.

With these two facts a manifold is a manifold regardless of:

which notion of neighbourhood you use
What you demand a neighbourhood is homeomorphic to

Definition of a manifold

Problem:

The open subset of $\mathbb{R}^n$ part

2 Answers2

For a topological manifold, $M$ with topology $\mathcal{J}$

I will write the proof in later, I've done one on paper

Linked