I Boundary and homeomorphism

[davidge]

The 2-sphere $\mathbb{S}^2$ can be expressed as the product $\mathbb{S}^1 \times \mathbb{S}^1$
Now can we express $\mathbb{S}^1$ as $\mathbb{S}^1 \subset (-a,a)$, where $(-a,a)$ is some open interval of $\mathbb{R}$? If so, then (I think) $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}$. Also, it's homemomorphic to $\mathbb{R}^2$ up to four coordinate charts covering it in $\mathbb{R}^2$.

If so, by the same reasoning $\mathbb{S}^2 \subset \mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$. Also, it's homeomorphic to $\mathbb{R}^3$, for we can define an embedding from it to $\mathbb{R}^3$.

Finally, if the above is correct, $\mathbb{S}^2 \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^3$, though it's not compact.

I'm trying to get this because I'm interested in knowing whether $\mathbb{S}^2 \times \mathbb{R}$ as a manifold has a boundary or not.

I'll appreciate any help.

 

[zwierz]

The 2-sphere S2\mathbb{S}^2 can be expressed as the product S1×S1

I was taught long ago that $S^1\times S^1$ is two dimensional torus not sphere.

 

[davidge]

I was taught long ago that $S^1\times S^1$ is two dimensional torus not sphere.

Oh, I thought the 2-sphere could be represented as $\mathbb{S}^1 \times \mathbb{S}^1$ because each one of the coordinates of a point $x \in \mathbb{S}^2$, say $(\theta, \phi)$, obeys a "circle equation".

 

[TeethWhitener]

Also, $\mathbb{S}^1$ is not homeomorphic to $\mathbb{R}$ (because it's not homeomorphic to an open interval of $\mathbb{R}$). Think about the values $\theta$ ranges over.

 

[davidge]

Also, $\mathbb{S}^1$ is not homeomorphic to $\mathbb{R}$ (because it's not homeomorphic to an open interval of $\mathbb{R}$). Think about the values $\theta$ ranges over.

So why can we map the entire circle using four coordinate charts from an open interval of $\mathbb{R}$? Can you check out my work below and say whether it's correct or not?

 

[fresh_42]

Oh, I thought the 2-sphere could be represented as $\mathbb{S}^1 \times \mathbb{S}^1$ because each one of the coordinates of a point $x \in \mathbb{S}^2$, say $(\theta, \phi)$, obeys a "circle equation".

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

 

[TeethWhitener]

So why can we map the entire circle using four coordinate charts from an open interval of $\mathbb{R}$? Can you check out my work below and say whether it's correct or not?

View attachment 204017

I can't really read what's going on here, but the easiest way to show that $\mathbb{S}^1$ is not homeomorphic to an open interval of $\mathbb{R}$ is to note that $\mathbb{S}^1$ is compact whereas an open interval of $\mathbb{R}$ is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

 

[davidge]

I can't really read what's going on here, but the easiest way to show that $\mathbb{S}^1$ is not homeomorphic to an open interval of $\mathbb{R}$ is to note that $\mathbb{S}^1$ is compact whereas an open interval of $\mathbb{R}$ is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

What if we choose a closed interval instead of a open one. A closed interval is compact.

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

Then letting the "end points", namely the poles out of consideration is forbidden?

 

[davidge]

closed interval has boundary, the circle does not

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to $\mathbb{R}^2$ using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to $\mathbb{R}$.

Can you point out where I'm wrong on the above ?

 

[fresh_42]

Then letting the "end points", namely the poles out of consideration is forbidden?

It's not forbidden, it simply makes the difference between local and global. As entire set $S^1 \times S^1 \ncong S^2$ as a torus (on the left) isn't a sphere (on the right), because one has a hole and the other has not. But on a small patch you can always apply coordinates like on earth. You simply cannot find a single chart describing both. As manifolds they all have local flat charts patched to a whole atlas, but this doesn't make them equal.

 

[TeethWhitener]

What if we choose a closed interval instead of a open one. A closed interval is compact.

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to R2R2\mathbb{R}^2 using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to RR\mathbb{R}.

Can you point out where I'm wrong on the above ?

I think you answered your own question. Look here:
https://www.physicsforums.com/threads/why-the-circle-cant-be-homeomorphic-to-a-real-interval.537897/
http://mathhelpforum.com/differenti...-not-homeomorphic-any-interval-real-line.html
Combining with what I said before: an open interval on $\mathbb{R}$ is not homeomorphic to $\mathbb{S}^1$ because compactness isn't preserved. A half-open interval won't work for the same reason. And as you said, there's no homeomorphism between a closed interval of $\mathbb{R}$ and $\mathbb{S}^1$. The links above answer why: choose a closed real interval $[a,b]$ and assume a (edit: bijection homeomorphism) exists $f: [a,b] \mapsto \mathbb{S}^1$. Then deleting any point $p$ (not equal to $a$ or $b$) from the real interval gives another homeomorphism: $g: [a,p) \cup (p,b] \mapsto \mathbb{S}^1 - \{f(p)\}$. But the new real interval $[a,p)\cup(p,b]$ is not connected, whereas $\mathbb{S}^1 - f(p)$ is connected. Since connectedness is a topological property, $g$ can't be a homeomorphism (and by extension, neither can $f$).

EDIT: one important note. The set $\mathbb{R} \cup \{\infty\}$ is homeomorphic to the circle. This is sometimes called the "point at infinity." It's a special case of one point compactification:
https://en.wikipedia.org/wiki/Alexandroff_extension

 

[davidge]

the new real interval $[a,p)\cup(p,b]$ is not connected

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval $(a,b)$, the same reasoning would led us to conclude that $\mathbb{S}^1$ and $(a,b)$ are not homeomorphic. Cool!

 

[zwierz]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

 

[WWGD]

The 2-sphere $\mathbb{S}^2$ can be expressed as the product $\mathbb{S}^1 \times \mathbb{S}^1$
Now can we express $\mathbb{S}^1$ as $\mathbb{S}^1 \subset (-a,a)$, where $(-a,a)$ is some open interval of $\mathbb{R}$? If so, then (I think) $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}$. Also, it's homemomorphic to $\mathbb{R}^2$ up to four coordinate charts covering it in $\mathbb{R}^2$.

If so, by the same reasoning $\mathbb{S}^2 \subset \mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$. Also, it's homeomorphic to $\mathbb{R}^3$, for we can define an embedding from it to $\mathbb{R}^3$.

Finally, if the above is correct, $\mathbb{S}^2 \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^3$, though it's not compact.

I'm trying to get this because I'm interested in knowing whether $\mathbb{S}^2 \times \mathbb{R}$ as a manifold has a boundary or not.

I'll appreciate any help.

There is a result, I think by Ulam, that $\mathbb S^n$ cannot be embedded in $\mathbb R^n$ -or-lower. Still, it would be interesting to see whether $\mathbb S^n$ " Is a square root" for some n , meaning whether it is homeomorphic to the product $Y \times Y$ of some topological space $Y$ . There is an argument showing $\mathbb R^{2n+1}$ is not a square root in this sense.

 

[WWGD]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

Nice. And there is also the connectivity one that any point p separates [a,b] , i.e., for any p in [a,b] , [a,b] -{p} is disconnected, while the same is not true for $\mathbb S^1$ - or higher-dimensions. I don't know the formal name for this, though ; it comes down to that if X is homeo to Y through h, then X-{p} is homeo to Y-{h(p)}. EDIT 2, besides, with heavier machinery, you also have contractibility, etc. which is preserved under homotopy equivalence alone. EDIT 3:

 

[fresh_42]

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

Nice idea. Or one can take $S^1 \times S^1 \cong \mathbb{P}^2(1,\mathbb{R}) \cong \mathbb{P}(2,\mathbb{R}) \ncong \mathbb{P}(1,\mathbb{C}) \cong S^2\;$, which indicates the problem with the poles.

 

[davidge]

there is a continuous function of $S^1$ to itself that does not have a fixed point.

what do you mean by having a fixed point?

 

[WWGD]

Yet another way :

what do you mean by having a fixed point?

It means there is a point that maps to itself, i.e., f(p)=p.

 

[davidge]

Yet another way :

It means there is a point that maps to itself, i.e., f(p)=p.

Thanks

 

[WWGD]

Thanks

No problem, consider re, Zwierz' post and fixed points the (continuous) map from $\mathbb S^1$ to itself that rotates every point by a fixed amount that is not an integer multiple of $2\pi$ .

 

[TeethWhitener]

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval $(a,b)$, the same reasoning would led us to conclude that $\mathbb{S}^1$ and $(a,b)$ are not homeomorphic. Cool!

You have to be careful here, because the topology on $S=[a,p)\cup(p,b]$ is inherited from $\mathbb{R}$, meaning that open sets on $S$ are defined as the intersection of an open set in $\mathbb{R}$ with $S$. So in this case, for example, $[a,p)$ would be an open set in the space $S$ (since it's the intersection of, e.g., $(a-1,p) \cap S$, where $(a-1,p)$ is an open set in $\mathbb{R}$), as would $(p,b]$. But of course, since $S-[a,p)=(p,b]$, this means that $S$ contains subsets that are both open and closed, and therefore $S$ cannot be connected.

 

[davidge]

You have to be careful here, because the topology on $S=[a,p)\cup(p,b]$

Why are you calling $S$ that way? I thought $S$ would be $[f(a), \ ... \ , f(p)) \cup (f(p), \ ... \ , f(b)]$ instead.

But of course, since $S-[a,p)=(p,b]$, this means that $S$ contains subsets that are both open and closed, and therefore $S$ cannot be connected.

I see

 

[TeethWhitener]

Why are you calling $S$ that way? I thought $S$ would be $[f(a), \ ... \ , f(p)) \cup (f(p), \ ... \ , f(b)]$ instead.

I see

Bad notational choice on my part. I meant $S$ as a subspace of $\mathbb{R}$, not a subspace of $\mathbb{S}^1$.

 

[davidge]

Just one more question: can we use the same argument to show that $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}^2$?

 

[fresh_42]

Just one more question: can we use the same argument to show that $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}^2$?

This cannot be shown because it is wrong.

 

[davidge]

This cannot be shown because it is wrong.

Since an embedding from $\mathbb{S}^1$ to $\mathbb{R}^2$ can be defined, I thought the former was homeomorphic to the later.

 

[fresh_42]

Homeomorphy means bijective and continuous in both directions. $\mathbb{S}^1$ is the unit circle. One can apply the stereographic projection from the northpole onto the equatorial plane and get a line $\mathbb{R}^1$ as a result. However, whereas the southpole can be seen as the origin, the northpole is the infinite point, and thus the circle is homeomorphic to the projective line. But there is no plane around, unless you do the same thing for the $2-$sphere.

The fastest way is still what @TeethWhitener said in post #7.

 

[WWGD]

Since an embedding from $\mathbb{S}^1$ to $\mathbb{R}^2$ can be defined, I thought the former was homeomorphic to the later.

You would require an embedding onto , not just into.

 

[davidge]

@fresh_42 and @WWGD
I think I see what you are saying.

You would require an embedding onto , not just into.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

 

[WWGD]

@fresh_42 and @WWGD
I think I see what you are saying.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

Yes, this is correct, but still does not show that the map is not onto $\mathbb R^2$, though. In this case you can also use some topological invariants to show the non-existence of a homeomorphism.

 

[fresh_42]

@fresh_42 and @WWGD
I think I see what you are saying.

Suppose a parametrization $g: (-r,r) \rightarrow \mathbb{S}^1 \\ t \mapsto x$ where $(-r,r)$ is an open interval of $\mathbb{R}$.
Now define $f_1: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, \sqrt{r²-x²} \bigg{)}$
This function does not cover the entire circle $\in \mathbb{R}^2$, because the points $(r,0)$ and $(-r,0)$ $\in \mathbb{R}^2$ are excluded. Also, we would need to define another function $f_2: \mathbb{S}^1 \rightarrow \mathbb{R}^2$ by $x = g(t) \mapsto \bigg{(}x, - \sqrt{r²-x²} \bigg{)}$

Is my above understanding correct?

Your notation is a bit confusing as I assume $f_1=g$ and $x=t$. A better parameterization is simply done be the angle. You can "run" through the circle by $p\, : \,[0,2\pi) \longrightarrow \mathbb{R}^2$ defined as $p(\varphi)=(\cos \varphi, \sin \varphi)$. This is a continuous embedding. But all points apart from the circle are not hit, so it's only injective, not surjective. A homeomorphism $h$ requires both, such that it can be inverted. In addition both mappings ($h$ and $h^{-1}$) have to be continuous.

$\mathbb{S}^1$ is a circle, $\mathbb{R}^2$ a plane. How could there be a bijection? So let's see what the stereographic projection does:

You can map all points smoothly from the circle to the line ($x-$axis): the southern half of the circle will get you the points between $-1$ and $+1$ and the northern half the points outside. Except for the northpole itself, which is infinitely far away. So you have to extend the real line by an infinitely far away point, which is called projective line (see link above). So now if the circle $\mathbb{S}^1$ and the projective line $\mathbb{P}(1,\mathbb{R})$ are topologically the same, it therefore can neither be the real line $\mathbb{R}^1$ nor the real plane $\mathbb{R}^2$.

 

[davidge]

you can also use some topological invariants to show the non-existence of a homeomorphism

Like the theorem that says that $A \times B$ is homeomorphic to $C$ iff both $A$ and $B$ are homeomorphic to $C$, or like the compactness property

Thanks @fresh_42 for the explanation. It is interesting to see in details how the stereographic projection works.

Now the problem is that if the circle is'nt homeomorphic to $\mathbb{R}^2$, then the definition of a boundary on a manifold doesn't seem to work in this case, because that definition says a manifold don't have a boundary if the neighborhood of each of its points is homeomorphic to $\mathbb{R}^n$. In the case of the circle, we know it does not have a boundary, but yet it's not homeomorphic to $\mathbb{R}^2$.

Or should we just use the fact that a manifold is required to be locally Euclidean and conclude that the circle is locally homeomorphic to $\mathbb{R}^2$? But if we proceeded this way, then we would find that there're no manifolds with boundaries at all and the whole concept of boundary would become meaningless.

 

[WWGD]

Actually circle is locally homeomorphic to $\mathbb R^1$ , not to $\mathbb R^2$. It is more like a twisted line than a solid object, informally.

 

[fresh_42]

Like the theorem that says that $A \times B$ is homeomorphic to $C$ iff both $A$ and $B$ are homeomorphic to $C$, or like the compactness property

What do you mean? If $A=B=C=\mathbb{S}^1$ is the circle, then $A \times B = \mathbb{S}^1 \times \mathbb{S}^1$ is a torus, which is not homeomorph to a circle $\mathbb{S}^1$ (wrong dimensions). Don't confuse homeomorphisms with homotopy mappings.

Now the problem is that if the circle is'nt homeomorphic to $\mathbb{R}^2$, then the definition of a boundary on a manifold doesn't seem to work in this case, because that definition says a manifold don't have a boundary if the neighborhood of each of its points is homeomorphic to $\mathbb{R}^n$. In the case of the circle, we know it does not have a boundary, but yet it's not homeomorphic to $\mathbb{R}^2$.

It gets automatically confusing here, as terms are used a bit ambiguously when it comes to manifolds compared to ordinary topological spaces. As manifolds are also topological spaces, this is a very unlucky situation, which in my language is solved by an extra name.
Let's see whether I can get it right.

The boundary $\partial S$ of a set $S$ in a topological space $X$ can be defined as $\partial S = \overline{S} \cap \overline{X - S}$.
As topological space, a manifold is automatically closed as well as open, which makes its boundary automatically empty. This is a direct result of the fact, that the manifold is the entire available space. No embeddings into some outer space!

As an embedded structure, like a ball in usual space, it has the the topology of this outer space and a boundary there: The open ball $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 < 1\}$ has the boundary $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 = 1\}$, the sphere, although it doesn't belong to the open ball anymore, or in case of $\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 \leq 1\}$ it has also the sphere as boundary which this time is part of the set, the closed ball. The space itself, here $\mathbb{R}^2$, has no boundary.

But as we don't want to have embeddings, when it comes to manifolds, we have to say goodbye to outer spaces.

Instead we have an atlas of our manifold $M$, which gives us the family of regular charts on open sets. Now to speak of a boundary, we consider the prototype of something with a boundary, namely the set $\mathbb{R}^n_+ = \{x \in \mathbb{R}^n\,\vert \,x_n \geq 0\}$. Next we allow besides our regular charts, some additional boundary charts for our sets $U \subseteq M$, that is a chart that is a 1-to-1 map $\phi$ such that $\phi(U) \subseteq \mathbb{R}^n_+$ is open. Regular points are those covered by regular charts, and boundary points are points which are not regular, that means covered by a boundary chart plus the condition $\phi(p)_n =0$. This way the boundary property of $\mathbb{R}^n_+$ is transported into terms of the manifold via its atlas.

So as for the circle, all remains as it is, i.e. no boundary at all since it has no interior.

However, if you mean the disc $M=\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 \leq 1\}$ instead, then the circle of this disc is its boundary in the sense that it contains the boundary points (charts which map to $\mathbb{R}^2_+=\{(x,y)\in \mathbb{R}^2\,\vert \,y\geq 0\}$).

If you take away these points, i.e. $M=\{(x,y) \in \mathbb{R}^2\,\vert \,x^2+y^2 < 1\}$, then it has no boundary points, as all points are covered by regular charts.

Or should we just use the fact that a manifold is required to be locally Euclidean and conclude that the circle is locally homeomorphic to $\mathbb{R}^2$?

The circle is a one dimensional curve and as such locally homeomorph to an open interval of the straight line $\mathbb{R}^1$.

But if we proceeded this way, then we would find that there're no manifolds with boundaries at all and the whole concept of boundary would become meaningless.

No, not really meaningless, just "handle with care". (See above, if I got it right. Pretty late here now.)
One last remark: The charts in an atlas don't have to be global, that is there is usually not a single chart that maps $M$ to $\mathbb{R}^n$, so there isn't a global homeomorphism needed. We only need local charts like in a real world atlas: pages of open sets which map to open sets in Euclidean space.

 

[davidge]

Thank you for the detailed reply.
Can you give an example of a map from the disc to $\mathbb{R}_{+}^{n}$?

So as for the circle, all remains as it is, i.e. no boundary at all since it has no interior.

Would another way to put it be to give a counter example e.g. the cylinder $[a,b] \times \mathbb{S}^1$ is not homeomorphic to $\mathbb{R}^2$ because $[a,b]$ and $\mathbb{S}^1$ are both compact, while $\mathbb{R}^2$ is not. So the cylinder is not locally (nor globally) homeomorphic to $\mathbb{R}^2$, which means it has a boundary.

No, not really meaningless, just "handle with care". (See above, if I got it right. Pretty late here now.)
One last remark: The charts in an atlas don't have to be global, that is there is usually not a single chart that maps $\mathbb{R}^n$, so there isn't a global homeomorphism needed. We only need local charts like in a real world atlas: pages of open sets which map to open sets in Euclidean space.

Yes, I see now

What do you mean? If $A=B=C=\mathbb{S}^1$ is the circle, then $A \times B = \mathbb{S}^1 \times \mathbb{S}^1$ is a torus, which is not homeomorph to a circle $\mathbb{S}^1$ (wrong dimensions). Don't confuse homeomorphisms with homotopy mappings.

Oh, I thought at first that there could be homeomorphisms between spaces of different dimensions.

 

[fresh_42]

Can you give an example of a map from the disc to $\mathbb{R}_{+}^{n}$?

Say $U$ is an open neighborhood of a point $(x,y)$ with $x^2+y^2=1$ on the closed disc $B$ as a manifold. We can chose a coordinate system, such that $(x,y)=(1,0)$ is the south pole of $B$ by an appropriate rotation and translation. Now the closed disc is completely within $\mathbb{R}_{+}^{n}$. Next we stretch $U$ in such a way, that all its boundary points come to rest on the boundary of $\mathbb{R}_{+}^{n}$, e.g. by a perspective mapping from the north pole.

The fraud here lies of course in the coordinate transformations $T$ where the work is done, but rotations as well as translations and the final stretching are continuous and bijective, i.e. a homoeomorphism.

This description is only to prevent me from doing any calculations, which probably won't show very much and the final stretching is like bending a curved piece of metal into a line, which is a bit difficult to write down. The imagination should do.

 

[lavinia]

You can see in many ways that $S^1×S^1$ is not homeomorphic to $S^2$. One way is to triangulate each and calculate the Euler characteristic from the triangulation. For the torus $S^1×S^1$ the answer is zero. For the sphere it is 2. This calculation is independent of the triangulation.

Also the sphere is simply connected while $S^1×S^1$ is not. This means that every closed loop on the sphere can be shrunk to a point - if one imagines that the loop is made of infinitely stretchable rubber - while the torus clearly has closed loops that can not be shrunk.

It is also true that every map of the sphere into $S^1×S^1$ is null homotopic - so it can not be a homeomorphism.

The sphere minus a point is contractible. That is: it can be continuously shrunk to a point. $S^1×S^1$ minus a point shrinks to a figure eight so it is not contractible.

 

[WWGD]

You can see in many ways that $S^1×S^1$ is no homeomorphic to $S^2$. One way is to triangulate each and calculate the Euler characteristic form the triangulation. For the torus $S^1×S^1$ the answer is zero. For the sphere it is 2. This calculation is independent of the triangulation.

Also the sphere is simply connected while $S^1×S^1$ is not. This means that every closed loop on the sphere can be shrunk to a point - if one imagines that it is made of infinitely stretchable rubber while the torus clearly has closed loops that can not be shrunk.

It is also true that every map of the sphere into $S^1×S^1$ is null homotopic - so it can not be a homeomorphism.

The sphere minus a point is contractible. That is: it can be continuously shrunk to a point. $S^1×S^1$ minus a point shrinks to a figure eight so it is not contractible.

I wonder if one can show that $S^2$ is also not a product space $Y \times Y$, nor even of the form $Y \times W$ EDIT: This may be false if we have $Y= S^2$ and Y={ $.pt$} , where {.pt} is a one-point space. but true otherwise.

. Is there any way to show $S^2$ is not the top space of a trivial bundle. I remember seeing an argument to the effect that $\mathbb R^{2n+1}$ is not a product space, i.e., it is not homeomorphic to the product $Y \times Y$ with the product topology..

 

[lavinia]

I wonder if one can show that $S^2$ is also not a product space $Y \times Y$, nor even of the form $Y \times W$ . Is there any way to show $S^2$ is not the top space of a trivial bundle. I remember seeing an argument to the effect that $\mathbb R^{2n+1}$ is not a product space, i.e., it is not homeomorphic to the product $Y \times Y$ with the product topology..

I am not sure. Let's think about it.

 

[zwierz]

another question: is the following implication true?: $\mathbb{R}^2=Y\times Y\Longrightarrow Y=\mathbb{R}$

 

[lavinia]

Thank you for the detailed reply.
Can you give an example of a map from the disc to $\mathbb{R}_{+}^{n}$?

For $n=2$ take a look at $z → (z + i)/(1 + iz)$ where $z$ is a point in the open unit radius disc centered at the origin.

 

[davidge]

For $n=2$ take a look at $z → (z + i)/(1 + iz)$ where $z$ is a point in the open unit radius disc centered at the origin.

But this is complex, not real.

 

[lavinia]

But this is complex, not real.

The complex plane is homeomorphic to $R^2$

 

[davidge]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

 

[WWGD]

another question: is the following implication true?: $\mathbb{R}^2=Y\times Y\Longrightarrow Y=\mathbb{R}$

Up to homeomorphism?

 

[WWGD]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

But a circle can be embedded in $\mathbb R^2$. The standard { $(x,y) \in \mathbb R^2 : x^2+y^2 =1$} is an embedding of the standard unit circle.

RE my previous post, Borsuk-Ulam theorem https://en.wikipedia.org/wiki/Borsuk–Ulam_theorem states that every continuous map f: $\mathbb S^n \rightarrow \mathbb R^n$ has a point x with f(x)=f(-x), meaning there can be no embedding of the n-circle into $\mathbb R^n$.

 

[fresh_42]

@fresh_42 I have something to add to our discussion regarding your post #34. I was reading some books on topology that are available at the university library and according to Differential Topology by Victor Guillemin and Allan Pollack: "The Whitney theorem states that every k-dimensional manifold embeds in $\mathbb{R}^{2k}$." Does that theorem implies that a homeomorphism (or as mentioned in that book, a diffeomorphism) can be defined from the k-dimensional manifold to $\mathbb{R}^{2k}$? If so, why did you say earlier that the circle (a 1-dimensional manifold) cannot be embedded in $\mathbb{R}^2$?

I haven't said it can't be embedded. I said it's of a different dimension and it can't be onto. There is no homeomorphism because there is no bijection. Every time you take a compass and draw a circle, you probably embed it in $\mathbb{R}^2$. And as a manifold, the total topological space is already merely the circle and there is no $\mathbb{R}^2$ where it lives in.

 

[davidge]

I said it's of a different dimension and it can't be onto

There is no homeomorphism because there is no bijection.

Ah, ok. Thank you

 

[davidge]

But a circle can be embedded in $\mathbb R^2$. The standard { $(x,y) \in \mathbb R^2 : x^2+y^2 =1$} is an embedding of the standard unit circle.

yes, I see. And can it also be embedded in $\mathbb{R}^p$, $p > 2$?

 

[zwierz]

Up to homeomorphism?

sure