# How do you prove that this function is bijective?

How do you prove that this function is bijective?
$f\colon (0,1)\longrightarrow \mathbb{R}$
$f(x)=\tan (\pi(x-1/2))$
In fact I want to show that $(0,1)$ is equivalent to $\mathbb{R}$ by proving that $f$ is bijective. Using Derivative concept and the Intermediate value theorem, it's easy to prove that is bijective, but I'm not allowed to use them.

• Well it follows from naive geometrical reasoning. Is that allowed? Commented Sep 4, 2015 at 20:58
• You could use the geometric properties of $\operatorname{tan}$. Commented Sep 4, 2015 at 20:58
• @RobArthan I just allowed to prove it by definitions of 1-1 and onto function, but I'd like to see your solution, too. Commented Sep 4, 2015 at 21:00
• What definitions of $\tan$ and $\pi$ are you using? Commented Sep 4, 2015 at 21:07
• Your "normally in calculus" is not my "normally in calculus", but let me put that to one side and put it another way: you've told us some properties of $\tan$ and $\pi$ that you are not allowed to use. What properties are you allowed to use? Commented Sep 4, 2015 at 21:21

"Draw" the unit half-circle with $x > 0$ and the tangent line at $(1, 0)$. Any point on that line determines a unique line segment to the origin, which interects the half-circle in a unique point . Conversely, each point on the half-circle uniquely determines a ray from the origin, which intersects the line in a unique point. Thus there is a bijection between the half-circle and the line. There is are bijections between the line and the real numbers and between the half-circle and the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, both based on directed distance from the point $(1, 0)$ along the respective curves. The combination of these three bijections is the function $\tan$. Therefore it is a bijection. Composing it with one other easily-seen bijection gives the result you want.