Using tan(x), show that open interval is diffeomorphic with the real line

Question

This is exercise 1.3 from An Introduction to Manifolds by Loring W. Tu

I want to show that $(a,b) \subset \mathbb{R}$ is diffeomorphic with the real line.

Strategy:

1.) I want to show that $$f:(-\frac{\pi}{2}, \frac{\pi}{2}) \rightarrow \mathbb{R}, f(x)=\tan(x)$$ is a diffeomorphism. How can I do that?

2.) Let $a,b$ two real numbers, $a<b$. How can I find a linear function $h:(a,b) \rightarrow (-1,1) $ With other words: How can I prove that any two finite intervals are diffeomorphic=. ?

3.) Can I conclude that the composition of $f$ and $g$ is a diffeomorphism of an open interval with $\mathbb{R} ?$

Edit: So my proof will be as follows:

$f(x)=\tan(x)$, where $f: (-\pi/2, \pi/2)\rightarrow \mathbb{R}$. $f$ is smooth, because the derivative $1/\cos^2(x)$ is smooth because $cos(x)$ is smooth and positive on the domain. Secondly $f^{-1}=\arctan(x)$ is a well defined function, which is equivalent by saying that f is a bijection. $\frac d{dx} \arctan(x) = 1/(1+x^2)$, so this function and all its derivatives are smooth $\rightarrow f^{-1}\in C^{\infty}$. Hence $f$ is a diffeomorphism.

Now $g:(a,b) \rightarrow (-1,1)$, where $g(x)=\frac 1{b-a}(x-a)-\frac1 2$ is a linear mapping, which is a bijection and both $g$ and $g^{-1}$ are smooth. $\implies$ two finite open sets are diffeomorphic.

Consider the composition $f \circ g: (a,b)\rightarrow \mathbb{R} $ is a diffeomorphism, hence $(a,b)$ and $\mathbb{R}$ are diffeomorphic.

Answer 1

For i) use that the arctan is the inverse function of the tan, both are monotone and the derivative of arctan is $$\frac{1}{1+x^2}$$ so it is diffferentiable.

for ii) you won't find a linear but an affine linear function, here a hint. Try to scale your intervall to the right diameter and say that $f(a)$ should be $-1$ and $f(b)=1$

iii) yes. There is a rule how to find the derivative of the composition of two functions.

Critique on your proof:

At first you didn't proof the bijection very well, as you just say that $f^{-1}$ is a smooth function, here someone could say that this is circular, as you don't know if $f^{-1}$ exists at all.

In general don't make life to hard, you only need $C^1$ not $C^\infty$, here it is the same work, so never mind.

For your function $g$, it is a monotone increasing function and $f(a)=-\frac{1}{2}$ and $f(b)=\frac{1}{2}$ so that is not a bijection.