Is tan: (-π/2, π/2) -> R a Homeomorphism?

[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

 

[Hurkyl]

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. )

 

[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?

 

[Hurkyl]

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)

 

[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.

 

[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).

 

[Nusc]

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

 

[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?

 

[Nusc]

Okay, is there anything wrong with it?