Homogeneous topological space

1. Nov 12, 2009

1. The problem statement, all variables and given/known data

For any $a \in \left( -1,1 \right)$ construct a homeomorphism $f_a: \left( -1,1 \right) \longrightarrow \left( -1,1 \right)$ such that $f_a\left( a \right) = 0$. Deduce that $\left( -1,1 \right)$ is homogeneous.

2. Relevant equations

Definition of a homogeneous topological space, ie that the exists a homeomorphism for each pair of points x,y which maps x to y.

3. The attempt at a solution

I can't find a set a functions which map an arbitrary point to zero and is surjective. My attemps include f = x - a, f = |x - a|, f = sin (x-a) but these are not homeomorphic for arbitrary a.

Last edited by a moderator: Nov 12, 2009
2. Nov 12, 2009

George Jones

Staff Emeritus
There are lots of such homeomorphisms. Examples:

1) join two straight lines that have different slopes (not necessarily equal to one);

2) use the points (-1, -1), (a, 0), and (1, 1) to pin down a parabola (I think this works, haven't worked the details).

3. Nov 12, 2009

LumenPlacidum

I'd be careful with the parabola idea, since if it actually has a peak in that interval, then it won't take open sets to open sets, but will take any open set that contains the x-value for the axis of symmetry to a half-open set.

4. Nov 12, 2009

George Jones

Staff Emeritus
Yes, that's why I said that I think it works. I drew some sketches that make it appear that the max/mins occur outside the interval, but I didn't solve any equations. My sketches could be misleading me.

Obvious choices of line segments in my example 1) give obvious homeomorphisms.

5. Nov 12, 2009

Can you explain what you mean by joining lines? Are you talking about intersecting lines in $$R^2$$?

6. Nov 12, 2009

George Jones

Staff Emeritus
Yes.

What are $A$ and $B$ such that $f_a \left( x \right) = Ax + B$ has

$$\lim_{x \rightarrow -1} f_a \left( x \right) = -1$$

and $f_a \left( a \right) = 0$?

Last edited: Nov 12, 2009
7. Nov 12, 2009

Right so you have a line starting at the point (-1,-1) and intersecting at (a,0). Unfortunately it does not finish at (1,1), and so it cannot be a bijection from the interval
(-1, 1) to (-1,1). Is this not right?

8. Nov 12, 2009

LumenPlacidum

It can be a bijection so long as the lines are both increasing or both decreasing.

9. Nov 12, 2009

Sorry I must be confused here. I can only see one line other than the interval (-1, 1). I thought you were talking about a line crossing the real axis at the point a.

10. Nov 12, 2009

George Jones

Staff Emeritus
So pick another line from (a,0) to (1,1), and turn f_a into a "schizophrenic" function.

11. Nov 12, 2009

ok so I have for x < a $$f_a \left( x \right) = \frac{x-a}{1+a}$$

and for x > a $$f_a \left( x \right) = \frac{x-a}{1-a}$$

Which coincide at x = a. Looks good to me.

12. Nov 12, 2009

George Jones

Staff Emeritus
Looks good if an equality is included for at least one of x < a, x > a.

The question is not done yet, though. You still need to show the deduction that gives that (-1, 1) is a homogeneous space.

13. Nov 12, 2009

I know I was just thinking about that. I need to show that for any x,y in the domain, there is a homeomorphism mapping x to y. Presumably I can somehow use the function I just made, but simple addition of y onto this function doesn't make a homeomorphism.

14. Nov 12, 2009

George Jones

Staff Emeritus
Unfortunately, I know only how to give a very small hint, or a very large hint (writing down the answer). Small hint: since both x and y are arbitrary elements of (-1 , 1), you should try things that have both x and y as indices.

Maybe someone else knows a better hint that doesn't give the whole answer away.

15. Nov 12, 2009

$$f_b \left( b \right) = 0, f_a^{-1} \left( 0 \right) = a, f_a^{-1} \left( f_b \left( b \right)\right) = a$$?

So that for any a,b we have a homeomorphism (i think) mapping b to a.

16. Nov 12, 2009

George Jones

Staff Emeritus
Yes, that's it.

I might switch the and b around in one of the places, i.e., either "So that for any b,a we have a homeomorphism," or $f_b^{-1} \left( f_a \left( a \right)\right) = b$.

17. Nov 12, 2009