# I Boundary and homeomorphism

1. May 21, 2017

### 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.

Last edited: May 21, 2017
2. May 22, 2017

### zwierz

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

Last edited: May 22, 2017
3. May 22, 2017

### davidge

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".

4. May 22, 2017

### 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.

5. May 22, 2017

### davidge

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?

6. May 22, 2017

### Staff: Mentor

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.

7. May 22, 2017

### TeethWhitener

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.

8. May 22, 2017

### davidge

What if we choose a closed interval instead of a open one. A closed interval is compact.
Then letting the "end points", namely the poles out of consideration is forbidden?

9. May 22, 2017

### davidge

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 ?

10. May 22, 2017

### Staff: Mentor

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.

11. May 22, 2017

### TeethWhitener

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

Last edited: May 22, 2017
12. May 22, 2017

### davidge

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!

13. May 22, 2017

### 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.

14. May 22, 2017

### WWGD

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.

Last edited: May 22, 2017
15. May 22, 2017

### WWGD

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:

Last edited: May 22, 2017
16. May 22, 2017

### Staff: Mentor

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.

Last edited: May 22, 2017
17. May 22, 2017

### davidge

what do you mean by having a fixed point?

18. May 22, 2017

### WWGD

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

19. May 22, 2017

### davidge

Thanks

20. May 22, 2017

### WWGD

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 $2\pi$ .

21. May 23, 2017

### TeethWhitener

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.

22. May 23, 2017

### davidge

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

23. May 23, 2017

### TeethWhitener

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

24. May 23, 2017

### davidge

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

25. May 23, 2017

### Staff: Mentor

This cannot be shown because it is wrong.