Constructing a Homeomorphism for Homogeneous Topological Spaces

[madness]

Homework Statement

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.

Homework 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.

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.

 

[George Jones]

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).

 

[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.

 

[George Jones]

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.

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.

 

[madness]

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

 

[George Jones]

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

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

 

[madness]

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?

 

[LumenPlacidum]

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

 

[madness]

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.

 

[George Jones]

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?

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

 

[madness]

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.

 

[George Jones]

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.

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.

 

[madness]

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.

 

[George Jones]

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.

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.

 

[madness]

How about this:

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

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

 

[George Jones]

How about this:

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

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

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

 

[madness]

Ok thanks that's the problem solved!