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## Primary tabs

# manifold

# Summary.

A manifold is a space that is locally like $\mathbb{R}^{n}$, however lacking a preferred system of coordinates. Furthermore, a manifold can have global topological properties, such as non-contractible loops, that distinguish it from the topologically trivial $\mathbb{R}^{n}$.

# Standard Definition.

An $n$-dimensional topological manifold $M$ is a second countable, Hausdorff
topological space^{1}^{1}For connected manifolds, the assumption that $M$ is
second-countable is logically equivalent to $M$ being paracompact, or
equivalently to $M$ being metrizable. The topological hypotheses in the definition of a manifold are needed
to exclude certain counter-intuitive pathologies. Standard
illustrations of these pathologies are given by the long line
(lack of paracompactness) and the forked line (points cannot be
separated). These pathologies are fully described in Spivak.
See this page.
that is locally homeomorphic to open subsets of
$\mathbb{R}^{n}$.

A differential manifold is a topological manifold with some additional
structure information. A *chart*, also known as a *system
of local coordinates*, is a mapping $\alpha:U\to\mathbb{R}^{n}$, such that the domain $U\subset M$ is an open set, and such that $U$ is homeomorphic to the image $\alpha(U)$. Let
$\alpha:U_{\alpha}\rightarrow\mathbb{R}^{n},$ and
$\beta:U_{\beta}\rightarrow\mathbb{R}^{n}$ be two charts with overlapping
domains. The continuous injection

$\beta\circ\alpha^{{-1}}:\alpha(U_{\alpha}\cap U_{\beta})\rightarrow\mathbb{R}^% {n}$ |

is called a *transition function*,
and also called a *a change of coordinates*. An atlas
$\mathcal{A}$ is a collection of charts $\alpha:U_{\alpha}\rightarrow\mathbb{R}^{n}$
whose domains cover $M$, i.e.

$M=\bigcup_{{\alpha}}U_{\alpha}.$ |

Note that each transition function is really just $n$ real-valued functions of $n$ real variables, and so we can ask whether these are continuously differentiable. The atlas $\mathcal{A}$ defines a differential structure on $M$, if every transition function is continuously differentiable.

More generally, for $k=1,2,\ldots,\infty,\omega$, the atlas $\mathcal{A}$ is said to define a $\mathcal{C}^{k}$ differential structure, and $M$ is said to be of class $\mathcal{C}^{k}$, if all the transition functions are $k$-times continuously differentiable, or real analytic in the case of $\mathcal{C}^{\omega}$. Two differential structures of class $\mathcal{C}^{k}$ on $M$ are said to be isomorphic if the union of the corresponding atlases is also a $\mathcal{C}^{k}$ atlas, i.e. if all the new transition functions arising from the merger of the two atlases remain of class $\mathcal{C}^{k}$. More generally, two $\mathcal{C}^{k}$ manifolds $M$ and $N$ are said to be diffeomorphic, i.e. have equivalent differential structure, if there exists a homeomorphism $\phi:M\to N$ such that the atlas of $M$ is equivalent to the atlas obtained as $\phi$-pullbacks of charts on $N$.

The atlas allows us to define differentiable mappings to and from a manifold. Let

$f:U\rightarrow\mathbb{R},\quad U\subset M$ |

be a continuous function. For each $\alpha\in\mathcal{A}$ we define

$f_{\alpha}:V\rightarrow\mathbb{R},\quad V\subset\mathbb{R}^{n},$ |

called the representation of $f$ relative to chart $\alpha$, as the suitably restricted composition

$f_{\alpha}=f\circ\alpha^{{-1}}.$ |

We judge $f$ to be differentiable if all the representations $f_{\alpha}$ are differentiable. A path

$\gamma:I\rightarrow M,\quad I\subset\mathbb{R}$ |

is judged to be differentiable, if for all differentiable functions $f$, the suitably restricted composition $f\circ\gamma$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. Finally, given manifolds $M,N$, we judge a continuous mapping $\phi:M\rightarrow N$ between them to be differentiable if for all differentiable functions $f$ on $N$, the suitably restricted composition $f\circ\phi$ is a differentiable function on $M$.

## Mathematics Subject Classification

53-00*no label found*57R50

*no label found*58A05

*no label found*58A07

*no label found*

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## Comments

## Closed manifolds vs boundaries?

Does the above definition say anything about the manifold being closed or not, or containing its boundary?

What about the open disk, the closed disk, and so on? Will they (with their natural topology) be regarded as 2-manifolds in the above sense? And what extra rule would we need to define the difference of being closed or not?

## Re: Closed manifolds vs boundaries?

> Does the above definition say anything about the manifold being closed

> or not, or containing its boundary?

Well manifold is topological space with additional structure, so as a topological space it is always closed. Also it has empty boundary (in a topological sense).

But if you think about classical definiton, ie. manifold is a subset of R^{n} with additional structure, then manifold does not have to be closed or open in R^{n}. For example set X={(x,0)| x<0} is a 1-dimensional manifold (without manifold boundary, but with topological boundary in R^{n}) embedded in R^{2} and it is neither open nor closed in R^{2}. On the other hand manifold X is closed as a topological space.

> What about the open disk, the closed disk, and so on?

Open disk in R^{n} is of course n-dimensional manifold (since it is homeomorphic to R^{n}). Closed disk is NOT a manifold in a classical sense (for example closed interval I should be a 1-dimensional manifold, but 1-dimensional connected manifolds are R^{1} and 1-dimensional sphere S^{1} - they are not homeomorphic to I).

Closed disk is a manifold with boundary. This is something more, then a manifold. And again boundary of manifold is not the same thing as topological boundary. But I think this is not the place to define what a manifold with boundary is. :)

Is a manifold with boundary always closed in R^{n}, after embedding? Again not! For example set

X={(x,y)| y>0, |x|<1} \cup {(x,0)| |x|<1}

is a 2-dimensional manifold with boundary, but it is neither open nor closed in R^{2}.

One thing is true - manifold with nonempty boundary cannot be open in R^{n}. This is trivial. :)

> And what extra rule would we need to define the difference of being

> closed or not?

Define what you mean by "closed" and I'll tell you whether such rule exist or not. ;] If you think about "closeness" in a sense I said before, then there is no such rule.

Tell me if there's anything else I can do to help you. :)

joking

## Re: Closed manifolds vs boundaries?

> Closed disk is a manifold with boundary.

Of course it is 2-dimensional manifold with boundary. :)

joking

## Re: Closed manifolds vs boundaries?

> Of course it is 2-dimensional manifold with boundary. :)

I meen closed disk in R^{n} is n-dimensional manifold with boundary. :) Sorry for trouble - it's pretty late here in Poland. :)

joking

## Re: Closed manifolds vs boundaries?

Thanks for your reply!!

I think my question was a little short-sighted - the answer should have been obvious to me, with a little more thinking. But maybe it is helpful to have that discussion down here, for others who have the same blindness getting the point.

Let's make this a more meaningful question.

The boundaries of a torus, a ball or any such shapes in RR^3 can't be modeled by homeomorphic transformations of the open disk, or the RR^2 plane - there would always be a stitching hole somewhere. So maybe the surfaces of these shapes share a property, that is missing in RR^2 or the open disk ?

Maybe the difference I'm looking for is just being compact?

## Re: Closed manifolds vs boundaries?

> The boundaries of a torus, a ball or any such shapes in RR^3 can't be

> modeled by homeomorphic transformations of the open disk.

Not true. Counterexample:

X={(x,y,z)\in R^{3} | x^2+y^2<1, z=>0 }.

This is 3-dimensional manifold with boundary embedded in R^{3}, but it's boundary (not topological but manifold) it's homeomorphic to open 2-dimensional disk.

As for torus - it is 2-dimensional manifold without boundary (unless you think about solid torus which is 3-dimensional manifold with boundary).

> So maybe the surfaces of these shapes share a property, that is

> missing in RR^2 or the open disk ?

That's what algebraic topology is all about. :) There are several ways to tell why for example torus is diffrent then R^{2}. And you're right that holes play special role in this movie.

The one way (and my favourite) is to look at homotopy groups. R^{n} has all homotopy groups trivial and it seems that the more topological space is "chaotic" the more nontrivial homotopy groups it has.

For example torus' fundamental group is ZxZ. Also n-dimensional's sphere's n-th homotopy group is Z. So not only these manifolds are not homeomorphic to R^{n} but they don't share the same properties (like homotopy type).

Of course compact spaces cannot be homeomorphic to any open subset of R^{n}, but they can have the same homotopy type (for example closed ball in R^{n} is a deformation retract of R^{n}). Therefore compactness does not tell us very much about spaces' "shape".

There are several diffrent ways like homology modules or Euler's characteristic.

Waiting for reply (if needed) :-)

joking

## Re: Closed manifolds vs boundaries?

Well, the boundary of X is like a drinking glass, that has a flat bottom, but is infinitely high. So I would say it is not "closed" in the same way as the surface of a ball, a torus, or a Klein bottle (not the surface, but itself, of course)

The mathematical difference I can see is for the boundary being compact. What I remember is, in a ball or torus surface, each sequence has an accumulation point (not true for your infinite drinking glass). PlanetMath has a another definition, but it also fails for the infglass. So maybe being compact is what I'm looking for?

## Re: Closed manifolds vs boundaries?

> Well, the boundary of X is like a drinking glass, that has a flat

> bottom, but is infinitely high.

You have to realise, that topological boundary and manifold boundary is not the same thing.

Topological boundary of X={(x,y,z)| x^2+y^2<1, z=>0} (as a subset of R^{3}) is

dX={(x,y,z)| x^2+y^2=1, z=>0} \cup {(x,y,0)| x^2+y^2<=1},

but manifold boundary is just

dX={(x,y,0)| x^2+y^2<=1}.

Really we cannot speak about topological boundary, because it all depends on topological space in which X is embedded. But manifold boundary depends only on manifold.

Also we cannot speak about closeness of manifold. It also depends on topological space in which manifold is embedded.

Remember that (manifold) boundary of a manifold with nonempty boundary is a manifold without boundary. :)

So if you talk about boundary of manifold being compact, you can just talk about compact manifolds without boundary.

So compactness of manifold is a way to tell diffrence between R^{n} and some manifolds. But there's nothing to be excited about. There cannot be homeomorphism between compact topological space and noncompact topological space (like R^{n}).

Also there are noncompact manifolds that are not homeomorphic (even don't have the same homotopy type) to R^{n} - such as X.

There's even more. There's a noncompact space with compact boundary. And it's not so difficult to figure it out. For example if B^{n} is an open ball in R^{n}, then R^{n}\B^{n} is noncompact manifold with nonempty compact boundary.

One more example: nonclosed (in R^{2}) manifold with compact boundary. This is simple. If D is a closed disk and B is another closed disk such that B and D have the same center, and radius of B is half of radius of D, then D\B is a manifold we are looking for. Manifold boundary of D\B is the same as topological boundary (in R^{2}) of D.

All I'm trying to say is that compactness is not really good. :) It's like trying to tell difference between dogs and other animals only looking at the number of legs. :)

joking

## Re: Closed manifolds vs boundaries?

ok..

Obviously I'm interested in the boundary (surface) of an object, as embedded in the euclidean space. Thus, the surface of your "half-infinite cylinder", as embedded in RR^3, has an "infinite drinking glass" as its boundary, which is homeomorphic to RR^2. This surface, in contrast to the surface of a ball, torus or whatever, is not compact.

The half-infinite cylinder itself is not really a manifold in the sense of the above definition, right? The points at the bottom don't have a sphere-shaped neighborhood in X.

Btw, is {(x,y)|x^2+y^2>1} homeomorphic to {(x,y)|x^2+y^2>0} ?

In this case, could we classify manifolds in those that are compact, and those that have one or more holes as above?

Sorry for this continuous asking :)

Actually, what I'm interested in is subdivided manifolds / boundary representations ("meshes"). A space X is subdivided into cells homeomorphic to a point, open interval, open disk, open 3-ball, k-ball, each cell 'surrounded' by boundary cells, and some more restrictions depending on the author. It seems a good idea to understand the underlying topological concepts.

For instance, things like the surface of a sphere can be modeled with a finite number of cells (cube map). The RR^2, or a ball with a point hole, needs an infinite number of cells (tiling with square-shaped cells..).

thx

## Re: Closed manifolds vs boundaries?

> Obviously I'm interested in the boundary (surface) of an object, as

> embedded in the euclidean space.

This can be tricky. Assume that M is manifold embedded in R^{2} (let say open ball). Then it has boundary (in our case 1-dimensional sphere). But on the other hand you can embedd it into R^{3} and boundary is diffrent (in our case entire M).

> The half-infinite cylinder itself is not really a manifold in the

> sense of the above definition, right? The points at the bottom don't

> have a sphere-shaped neighborhood in X.

True. This is not a manifold in classical sense.

> Btw, is {(x,y)|x^2+y^2>1} homeomorphic to {(x,y)|x^2+y^2>0} ?

Yes, it is. Let A(x,y)=|(x,y)|-1, where |(x,y)| is an Euclidean norm. Then you have function:

f(x,y)=(A(x,y)*x, A(x,y)*y),

which is homeo (or I just hope so - sorry, I'm to lazy to check ;] ).

> In this case, could we classify manifolds in those that are compact,

> and those that have one or more holes as above?

And how will you define holes? ;)

> A space X is subdivided into cells homeomorphic to a point, open

> interval, open disk, open 3-ball, k-ball, each cell 'surrounded' by

> boundary cells, and some more restrictions depending on the author.

> It seems a good idea to understand the underlying topological

> concepts.

This sounds like a CW complex or even regular CW complex. Maybe CW complexes are things you should focus on?

joking

## Re: Closed manifolds vs boundaries?

> This sounds like a CW complex or even regular CW complex.

> Maybe CW complexes are things you should focus on?

I just had a look at that bitch, here in planetmath - why can't you math guys make a common sense definition?? I totally miss the clarifying word "partition" - or is it me who is wrong? I think Brisson has a nice definition, but calls it "subdivided manifold", see http://portal.acm.org/citation.cfm?id=73858&coll=&dl=ACM&CFID=15151515&C...

Ok, but I should probably discuss that on the CW-complex page..

## Re: Closed manifolds vs boundaries?

> And how will you define holes? ;)

Well, for a manifold X of dim n, I would try to identify a subset S of X homeomorphic to the closed (!) n-ball with a point hole in the middle. If we are right about the homeoness of holes, this would let us identify a hole in the infinity of RR^n, at the outer limit of the open n-ball, at the end of an infinite cylinder mantle, etc.

Maybe there can be other, more freaky holes, though..

## Re: Closed manifolds vs boundaries?

> I would try to identify a subset S of X homeomorphic to the closed (!)

> n-ball with a point hole in the middle.

Hmmm... what about 1-dimensional sphere (circle)? Or something like this:

X={(x,y,z)| x^2+y^2=1; -1<z<1}?

Does X or circle have a hole? Does not seem to have your hole. :)

This entire "hole" problem is not as easy as it may look like.

joking

## Re: Closed manifolds vs boundaries?

The posts are getting thinner... (need to change the indentation)

I need to be more precise, one thing was missing.

If the surface of an n-ball or a torus is embedded (in a homeo way) into any space Y, it will still be a closed subset of Y (Y without f(X) will be open in Y).

If you embed the RR^n, an open ball or anything into another space, then it can easily happen that it's not a closed subset of this other space.

Now let X be the n-dim manifold we're interested in.

My idea is to identify a closed subset S of X (S is closed in S, iff X\S is open in the topology of S), which is homeomorph to the closed n-ball with a point hole inside. By identifying all such subsets of X, we can tell how many holes there are in X. Each hole has the same "dimension" as X.

More precisely, we are looking for a collection H of subsets of X, such that

(i) the elements of H don't intersect

(ii) each S in H is closed in X

(iii) each S in H is homeomorphic to the closed n-ball with a point hole.

(iv) "completeness": There is no further subset of X that could be added to H.

The set H identifies the holes in X.

> Hmmm... what about 1-dimensional sphere (circle)?

A 1-ball is an open interval between -1 and 1, right? If a sphere is the boundary of a ball, then a 1-sphere is the set {-1, 1}. Which is quite untypical for a sphere, because it's not connected.

The 2-ball would be an open disk, the 2-sphere would be the looped line around.

In the sense of above, any n-ball with n>1 has exactly one hole, as identified by the set {(x0...xn}|0.5 <= x0Â²+...+xnÂ² < 1}, which is homeomorphic to the closed n-ball with a point hole. ('hollow earth' transformation)

It's tricky for the 1-ball. The definition is clear, but maybe needs a change for dim 1. In a new definition, the interval [0,1) would have one hole, the interval (-1, 1), which is the open 1-ball, would have two.

Any of the n-spheres (=surface of the n-ball) have no holes. Embedded in RR^n (or any other space), they remain closed subsets.

> X={(x,y,z)| x^2+y^2=1; -1<z<1}?

A 2-dim manifold with two holes.

The holes can be identified as the subsets

A = {(x,y,z)| x^2+y^2=1; -1<z<=-0.5} and

B = {(x,y,z)| x^2+y^2=1; 0.5<=z<1},

which are both closed in X, but homeomorphic to the closed 2-ball with a point hole.

So far, my concept is proving well :)

> Does X or circle have a hole? Does not seem to have your hole. :)

This depends on my question some posts before. Is a point hole homeomorphic to a bigger hole? Is a (n+1)ball surface with a point hole homeomorphic to the RR^n ?

If yes (which seems so), all holes should be the same.

If not, we could go on to classify the holes of a manifold.

## Re: Closed manifolds vs boundaries?

> then a 1-sphere is the set {-1, 1}

Just to be precise: there's a convention that n-sphere is a boundary of n+1-dimensional closed disk. :)

Still, there's no hole (in your sense) in 1-sphere. Why?

Assume there's a hole (in your sense). Since there's hole, then there's subset of 1-sphere homeomorphic to 1-dimensional closed ball without point (there's no other possibilities, since 1-sphere is 1 dimensional manifold, so you cannot embed in 1-sphere any n-dimensional sets, like n-ball without point, for n>1). So then we have (for example) homeomorphism (onto image)

f:[-1,0) \cup (0,1]->S^{1},

because [-1,0) \cup (0,1] is exactly 1-dimensional closed disk without point.

Since it is homeomorphism (onto image), then in particular this is open map. Since [-1,0)\cup (0,1] is open (as a topological space), then image of f is open in S^{1}. Also it cannot be entire S^{1} (because image is not connected) so it is not closed. Contradiction. There are no holes in S^{1}.

I can say even more - this can be easily generalized to statement: "there are no holes in n-sphere for any n".

> A 2-dim manifold with two holes.

> The holes can be identified as the subsets

> A = {(x,y,z)| x^2+y^2=1; -1<z<=-0.5} and

> B = {(x,y,z)| x^2+y^2=1; 0.5<=z<1},

> which are both closed in X, but homeomorphic to the closed 2-ball

> with a point hole.

>

> So far, my concept is proving well :)

I'm not sure. :) This manifold is just a kind of "pipe". There should be no more then 1 hole (unless it is broken pipe ;] ).

There's more. Open ball in R^n has a hole (in your sense)!!! ;]

Assume B is the n-dimensional open ball in R^n. We can assume that:

B={(x1,...,xn)| x1^2+...+xn^2<2}.

Let C={(x1,...,xn)| 1<=x1^2+...+xn^2<2}.

It is clear that C is closed subset of B. Also realize that C is homeomorphic to n-dimensional ball without point. I'll even give this homeo. ;) Recall that C'={(x1,...,xn)| 1/2<x1^2+...+xn^2<=1} is homeomorphic to closed n-ball without point (talked about it few posts ago). Define homeo:

f:C->C'

f(x):=(1/|x|)*x.

It is inverse to itself, so this is homeo.

> Is a point hole homeomorphic to a bigger hole? Is a (n+1)ball

> surface with a point hole homeomorphic to the RR^n ?

Yes. Yes. :)

> If yes (which seems so), all holes should be the same.

I don't think so. I think there's a diffrence between hole in 1-sphere and hole in 2-sphere. Also torus has 2 holes which are not the same (we could say: one is inside and one is outside).

Sorry to disappoint you, but still you're doing great job. :) Keep on trying, never give up - that's how great minds were born. ;)

joking

## Re: Closed manifolds vs boundaries?

Hehe, you should be more careful when reading my posts :)

I think you missed some points - there is less we disagree about than you seem to think.

=- Just to be precise: there's a convention that n-sphere is a boundary of n+1-dimensional closed disk. :) -=

Ok, then 0-sphere is two points, while 1-sphere is the round line around a circle.

=- there's no hole (in your sense) in 1-sphere. -=

That's what I'm trying to say. No hole in a sphere.

I want a hole to be understood as something that prevents the space from being compact, and from being embedded into other (compact) spaces as a closed subset. These are places where the space is missing something - an explicit boundary.

I'm not talking about holes as the one in a pipe, or a torus.

=- I can say even more - this can be easily generalized to statement: "there are no holes in n-sphere for any n". -=

see my previous post :)

=~- Any of the n-spheres (=surface of the n-ball) have no holes. Embedded in RR^n (or any other space), they remain closed subsets. -~=

=- I'm not sure. :) This manifold is just a kind of "pipe". There should be no more then 1 hole (unless it is broken pipe ;] ) -=

It's not about being a pipe or something. What's important is the "z<1" vs "z<=1". At this end, a lot of sequences lack their accumulation points.

So, there's one hole for z near 1, and one for z near -1.

=- Open ball in R^n has a hole (in your sense)!!! -=

Yes, it has one hole, because it's missing exactly one connected component of boundary when embedded into a compact space.

=- I don't think so. I think there's a diffrence between hole in 1-sphere and hole in 2-sphere. Also torus has 2 holes which are not the same (we could say: one is inside and one is outside). -=

Yes, n-dim manifolds have n-dim holes. Well, a point hole is always 0-dim, but the piece of space around is n-dim.

Now, the easy model actually breaks with the torus, and similar structures - it seems, there can be more complicated holes for manifolds of dim 3 or higher.

The inner of a torus (as a 3-manifold, a volume) has exactly one hole - if embedded in RR^3, we need one connected component of boundary to make it a closed subset of RR^3. If embedded into a more obscure 3-manifold, it will be sufficient to add a circle (1-sphere, right?) to make it a closed subset.

The points in the torus can be specified by the angles alpha and beta (each from 0 to 2pi), and one radius coordinate r in [0,1). We do already say that two points with r1=r2=0 and alpha1=alpha2 are the same, no matter if beta1!=beta2. Now, to extend the torus to a compact 3-manifold, we allow r to be 1, and say if r1=r2=1, and alpha1=alpha2, then the two points are equal, no matter for the betas.

However, we can't "repair" the torus by adding a simple point, as this point would not have a 3-manifold neighbourhood.

## Re: Closed manifolds vs boundaries?

Hm, it's even easier with the torus.

An open n-ball can be turned into a compact n-manifold by adding a single point ('hollow earth' transformation). I would guess this will turn it into an n-sphere, but I'm not sure for higher dimensions.

A torus (the inner of it) is homeomorphic to the cross product of a 1-sphere (= looped line) and an open disk (2-ball). So we just need to add one point to turn the non-compact open disk into a compact 2-sphere. This point is turned into a 1-sphere when building the cross product again. So the thing we need to add is a 1-sphere, to make the torus a compact 3-manifold.

There are sometimes different alternatives to make a manifold compact - but we are always looking for an addition of lowest dimension.

For non-orientable manifolds it gets more tricky to imagine.

I would guess a moebius strip needs one extra point to make it a Klein bottle (which is compact) - but again I'm not sure.

Now consider the cross product of a Klein bottle and an open interval (1-ball). Of course we can add a single point to the 1-ball, which turns it into a 1-shere, and then build the cross product - which means, we add a complete Klein bottle (dim 2). But can't we add something of lower dimension? A point, or a 1-sphere?

We could also consider the cross product of a Klein bottle and higher dim n-balls, or the non-orientable spaces from projective geometry..

Tricky, tricky - but looks interesting to me :)

## Re: Closed manifolds vs boundaries?

Ohhh... sorry - missed that post. :)

I think now I understand your idea. It seems to work for manifold, since it is "good" space. Still it's strange to use word "hole". Intuition says that there is hole in a sphere (at least 1-dimensional) or in surface of torus (there's hole in "inner torus" but there's no hole in surface?). So maybe you should use diffrent word. :)

> However, we can't "repair" the torus by adding a simple point, as this

> point would not have a 3-manifold neighbourhood.

Why? Due to Alexandrov theorem every Hausdorff, locally compact (but not compact) space can be embedded in one-point compactification. It is unclear to me why this compactification won't be a manifold, when base space is manifold (for example compactification of n-dimensional real space is n-dimensional sphere). I have this strange feeling that it is manifold - I'll think about it.

joking

## Re: Closed manifolds vs boundaries?

> Why? Due to Alexandrov theorem every Hausdorff, locally

> compact (but not compact) space can be embedded in one-point

> compactification. It is unclear to me why this

> compactification won't be a manifold, when base space is

> manifold (for example compactification of n-dimensional real

> space is n-dimensional sphere). I have this strange feeling

> that it is manifold - I'll think about it.

Unfortunately not. The 1-point compactification of the annulus is not a manifold, for example.:

There is a base of open sets at the new point that are all homeomorphic to two 2-balls joined at a single point.

No connected open set in R^2 can be disconnected by removing a single point like this, so no open set in the 1-point compactification of the annulus is homeomorphic to an open set in R^2.

Hence the 1-point compactification of the annulus is not a manifold.

## Re: Closed manifolds vs boundaries?

Another example is two, open, disjoint intervals. It is 1-dimensional manifold, but it's compactification is the wedge sum of two circles, which is not a manifold.