Show that a "cross" is not a topological manifold

[cianfa72]

TL;DR

How to show that a "cross" in the plane is not a topological manifold for any n.

Hi, I've a doubt about the following example in "Introduction to Manifold" by L. Tu.

My understanding is that if one assumes the subspace topology from $\mathbb R^2$ for the "cross", then one can show that the topological space one gets is Hausdorff, second countable but non locally homeomorphic to any $\mathbb R^n$.

However how can one show that there is not any other topology on the "cross" such that its intersection point $p$ is homeomorphic to some $\mathbb R^n, n \geq 1$ ?

Thank you.

 

[fresh_42]

If $(X_1,\mathcal{T_1})\stackrel{f}{\longrightarrow }(X_2,\mathcal{T_2})$ is a homeomorphism, then $\mathcal{T_1}=f^{-1}\left(\mathcal{T_2}\right).$ Hence, if the cross is given the subspace topology of $\mathbb{R}^2$ to define $\mathcal{T_2}$ as it is the assumption of the exercise, you don't have any choice for $\mathcal{T_1}$ anymore.

 

[cianfa72]

If $(X_1,\mathcal{T_1})\stackrel{f}{\longrightarrow }(X_2,\mathcal{T_2})$ is a homeomorphism, then $\mathcal{T_1}=f^{-1}\left(\mathcal{T_2}\right).$ Hence, if the cross is given the subspace topology of $\mathbb{R}^2$ to define $\mathcal{T_2}$ as it is the assumption of the exercise, you don't have any choice for $\mathcal{T_1}$ anymore.

Yes, as far as the exercise goes. However my point is: suppose there was a bijection $f$ between a neighborhood of the intersection point $p$ in the "cross" and an open ball in $\mathbb R^n$ for some $n \geq 1$. Then one could define the corresponding topology on that neighborhood via $f^{-1}$ establishing an homeomorphism (basically by definition).

How can one show that the above is not the case ?

 

[jbergman]

TL;DR Summary: How to show that a "cross" in the plane is not a topological manifold for any n.

Hi, I've a doubt about the following example in "Introduction to Manifold" by L. Tu.
View attachment 349156

My understanding is that if one assumes the subspace topology from $\mathbb R^2$ for the "cross", then one can show that the topological space one gets is Hausdorff, second countable but non locally homeomorphic to any $\mathbb R^n$.

However how can one show that there is not any other topology on the "cross" such that its intersection point $p$ is homeomorphic to some $\mathbb R^n, n \geq 1$ ?

Thank you.

If you put a different topology on it, it isn't the cross any more. Without a topology it is just a set of points. The topology determines which points are near which and in what manner. If I put the discrete topology on this set, every point is equally near each other point.

 

[cianfa72]

If you put a different topology on it, it isn't the cross any more. Without a topology it is just a set of points. The topology determines which points are near which and in what manner. If I put the discrete topology on this set, every point is equally near each other point.

Ah ok, so basically "the cross" implicitly assumes (let me say by definition) the subspace topology from standard $\mathbb R^2$ topology.

I'm often confused by "intrinsic" vs "extrinsic" definition of topology on a set of points. In this case "the cross" is topologized in an extrinsic fashion.

 

[pasmith]

You can also define the cross as a quotient X = (\mathbb{R} \times \{0,1\}) / \sim where \sim identifies (0,0) and (0,1). The topology on \mathbb{R} is the standard topology and the topology on \{0,1\} is discrete.

 

[cianfa72]

You can also define the cross as a quotient X = (\mathbb{R} \times \{0,1\}) / \sim where \sim identifies (0,0) and (0,1). The topology on \mathbb{R} is the standard topology and the topology on \{0,1\} is discrete.

Ah ok, so according this viewpoint "the cross" topology is no longer extrinsically defined.

 

[WWGD]

Well, you have the connectedness number, I think some Euler-named invariant. Removal of the central point in the cross disconnects it , in the standard topology, into 4 parts. No such thing with the removal of one point from an open ball/set in $\mathbb R^2$.

 

[WWGD]

Basically, if there's a homeomorphism $h$ between the cross C and an open set U of Euclidean space, let $p$ be the central point. Then use any homeomorphism restricts to a homeomorphism( Careful! Converse is false!). Then h': =h|C-{p}--> U-{h(p)} is a homeomorphism between the cross without its center, and an open ball without one point. The first is disconnected, the latter isn't.

 

[cianfa72]

Well, you have the connectedness number, I think some Euler-named invariant. Removal of the central point in the cross disconnects it, in the standard topology, into 4 parts. No such thing with the removal of one point from an open ball/set in $\mathbb R^2$.

Yes, that is (in bold) the point: "the cross" topological space is defined using the subspace topology from $\mathbb R^2$ standard topology.

 

[WWGD]

Yes, that is (in bold) the point: "the cross" topological space is defined using the subspace topology from $\mathbb R^2$ standard topology.

Yes, and the result holds under the subspace topology, showing it isn't homeomorphic to the Euclidean 2-space.

 

[Office_Shredder]

In fact if you get to attach an arbitrary topology to that set of points, you don't have to do much finagling to get a manifold. Separate the cross into four lines, one of which arbitrarily gets the center point. You have three lines with one endpoint and one with both endpoints. Glue them in the obvious way and you get a line with endpoints which is a manifold with a boundary.

Of course the trick here is that under this topology there are neighborhoods of that center point that don't include all four parts of the cross, which is why the connectivity thing falls apart.

 

[cianfa72]

In fact if you get to attach an arbitrary topology to that set of points, you don't have to do much finagling to get a manifold. Separate the cross into four lines, one of which arbitrarily gets the center point. You have three lines with one endpoint and one with both endpoints. Glue them in the obvious way and you get a line with endpoints which is a manifold with a boundary.

Sorry, in which way one must glue those 3 + 1 parts in order to get a line with endpoints included ?

Of course the trick here is that under this topology there are neighborhoods of that center point that don't include all four parts of the cross, which is why the connectivity thing falls apart.

How is defined the topology you were talking about: is it the subspace topology from $\mathbb R^2$ on the closed line one gets as above (i.e. manifold with a boundary) ?

 

[Office_Shredder]

Sorry, in which way one must glue those 3 + 1 parts in order to get a line with endpoints included ?

You have to glue the open ends to closed ends, at which point every choice you make is find e

How is defined the topology you were talking about: is it the subspace topology from $\mathbb R^2$ on the closed line one gets as above (i.e. manifold with a boundary) ?

The topology I defined has nothing to do with the subspace topology on the cross. I was just giving a different choice of topology on the set of points that form a cross to demonstrate that the choice of topology is the thing that gives you a disconnected set when you remove a point in the middle, and it's not an inherent property of the set of points themselves

 

[cianfa72]

You have to glue the open ends to closed ends

Ok, as in the following picture I drew

The topology I defined has nothing to do with the subspace topology on the cross. I was just giving a different choice of topology on the set of points that form a cross to demonstrate that the choice of topology is the thing that gives you a disconnected set when you remove a point in the middle, and it's not an inherent property of the set of points themselves

Yes, that was my point too. My claim was that the topology on the set on the closed segment on the right is the subspace topology from $\mathbb R^2$. With this topology, removing the point A in the topological space defined this way, no longer disconnect it.

 

[Office_Shredder]

Ok, as in the following picture I drew
View attachment 349618


Yes, that was my point too. My claim was that the topology on the set on the closed segment on the right is the subspace topology from $\mathbb R^2$. With this topology, removing the point A in the topological space defined this way, no longer disconnect it.

There is a set bijection between the set of points that form a cross and the set of points that form a closed interval. There is no homeomorphism between them using the subspace topology each receives considered as subsets of $\mathbb{R}^2$.

The gluing procedure I described is not a continuous function under the subspace topology, so there's no reason it would preserve connectedness. So yes, it turns out under this kind of weird topology you can make the cross connected

 

[cianfa72]

There is a set bijection between the set of points that form a cross and the set of points that form a closed interval. There is no homeomorphism between them using the subspace topology each receives considered as subsets of $\mathbb{R}^2$.

Ok, right.

The gluing procedure I described is not a continuous function under the subspace topology, so there's no reason it would preserve connectedness. So yes, it turns out under this kind of weird topology you can make the cross connected.

Yes, so basically the topology on "the cross" as set is by construction the same as the topology on the closed interval (i.e. by the set bijection we define/transport the closed interval topology to "the cross" set which then becomes an homeomorphism by definition).

 

[Office_Shredder]

That's right

 

[WWGD]

Yes; if S is a set and X a topological space, f a function, just define, for $U \in \tau_X:f^{-1}(U)$ open in S.

Not sure if Musk has given X a topology , though ;).

 

[cianfa72]

Not sure if Musk has given X a topology , though ;).

Sorry, what is Musk ?

 

[WWGD]

Sorry, what is Musk ?

Elon Musk; from X, Tesla.