# Sin t = 0 homeomorphism

1. Sep 23, 2006

### Nusc

Show that tan: (-pie/2,pie/2)->R is a homeomorphism where tan = sin/cos

To show that f and f^-1 are cts, it seems trivial from a sketch but how do you do it?

For 1-1 tan(x) = tan(y)

Need to knwo x =y

tan(x) = sinx.cosx = siny/cosy = tany

=> sixcosy = sinycosx

this gets you sin(x-y) = 0
But x-y = pie

What's wrong here?

Onto is obvious

2. Sep 23, 2006

### Hurkyl

Staff Emeritus
There are infinitely many solutions to the equation sin t = 0... but not many that are consistent with the conditions of your problem.

(Incidentally, it's not just a homeomorphism, but also a diffeomorphism. That's easier to prove, since we're in dimension 1. )

3. Sep 23, 2006

### Nusc

Then it must be 0 because the domain is the interval (-pie/2,pie/2)

So how do I show that f and f^-1 are cts?

4. Sep 24, 2006

### Hurkyl

Staff Emeritus
How have you defined them? e.g. showing tan to be continuous is trivial if you've defined sine and cosine via power series (or are allowed to use their power series)

5. Oct 9, 2006

### Nusc

So for onto
for all y in R there exists an x in (-pie/2,pie/2) s.t. f(x)=y

WTS: tanx = y

Let y = siny/cosy = tany

let x = y => y = tany = tanx

is that sufficient? y could be inf.

6. Oct 9, 2006

### matt grime

How can y be 'inf'? Neither is infinity in R, nor is pi/2 (pi does not have an e) in the interval (-pi/2,pi/2).

7. Oct 9, 2006

### Nusc

So then how would you show f^-1 is cts ?

8. Oct 16, 2006

### Nusc

Okay let me try again.

To show that f is cts pick an open subset G in R. then f-1 = arc tanx: R -> -pie/2, pie/2)

Suppose G is the interval (a,b).

Let f-1(G) = A. Thus A is contained in (-pie/2,pie/2)

f-1(G) is an element of (f-1(a),f-1(b))=A

which is an open subset and therefore cts.

To show that the inverse of f is cts.

we want to show that for any open G in (-pie/2,pie/2) thus f-1(G) = A where A is open.

IF G = (a,b) then f-1(G) is an element of (f-1(a),f-1(b)) = (tana,tanb) which is an open interval.

and hence cts.

Is this correct?

9. Oct 17, 2006

### Nusc

Okay, is there anything wrong with it?