Does a bijective map from $(-\pi/2,\pi/2)\to (0,1)$ exist? 
Does a bijective map from $(-\pi/2,\pi/2)\to (0,1)$ exist?

My first guess was using the sine function but it doesn't really comply with the bijective map since it goes from  $(-\pi/2,\pi/2)\to (-1,1)$. Am I missing something with the sine function or is there another way I can achieve this bijection? 
 A: Consider the function $f(x) = \frac{x}{\pi} + \frac{1}{2}$.
A: As @leibnewtz said, for any two intervals $(a,b), (c,d)$ there is a function as you asked. Consider 
$$f(x)=\frac{x-a}{b-a}(d-c)+c.$$
In this case, 
$$f(x) = \frac{x-(-\pi/2)}{\pi/2-(-\pi/2)}(1-0)+0 = \frac{ x + \pi/2}{\pi}=\frac{x}{\pi}+\frac{1}{2}$$
as @qaphla has pointed out. 
Note that $f$ is continuous, $f(a)=c, f(b)=d$ and $f'(x)=\frac{d-c}{b-a}>0$, hence bijective. 
A: Take any bijection $f(x)$ from $(0,1)$ to $(0,1)$. Take a linear change of variable like $l_{-}(y) = \frac{1-2y}{2\pi} $ or $l_{+}(y) = \frac{1+2y}{2\pi} $. Then $f(l_\pm(y))$ is a bijection from $(-\frac{\pi}{2},\frac{\pi}{2})$ to $(0,1)$, because $l_{-}$ or $l_{+}$ maps  $(-\frac{\pi}{2},\frac{\pi}{2})$ to $(0,1)$ first, and is bijective. 
Now for $f(x)$ you can take for instance $x\to x^p$ or $x\to \sin^p(2x/\pi)$, with  $p>0$, and you have at least a bijection with a sine function.
A: Any interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to any other interval $(c,d)$, which in particular implies that there is a bijection $f:(a,b) \to (c,d)$.
