# 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?

• Notice that the diameter of both intervals are the same, so it's really easy to come up with a bijective function in this case (as qalpha has already done below). – Irregular User May 29 '16 at 20:16

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

• $f(-\pi/2)=0$ and the function is increasing… – egreg May 29 '16 at 20:24
• @egreg Yes, and the question is asking for a bijection from $(\frac{-\pi}{2}, \frac{\pi}{2})$ to $(0, 1)$. What's the problem? – qaphla May 29 '16 at 20:30
• Sorry, I was misled by the text of the question and didn't read carefully the title. – egreg May 29 '16 at 20:34

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.

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.

Any interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to any other interval $(c,d)$, which in particular implies that there is a bijection $f:(a,b) \to (c,d)$.

• How do you prove the homeomorphism? By constructing a bicontinuous bijection. So this is no answer. – egreg May 29 '16 at 20:25
• The bicontinuous bijection may be constructed without reference to specific values of $a$, $b$, $c$, and $d$. The question was if a bijection exists, and not for a specific example of such a bijection – leibnewtz May 29 '16 at 20:28
• Let $f(x)=\frac{x-a}{b-a}(d-c)+c$. – B. Rivas May 29 '16 at 20:48