real analysis - 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?

Answer

Consider the function $f(x) = \frac{x}{\pi} + \frac{1}{2}$.