# Is there a function from $(0, 1)$ to $\mathbb{R}$ that is surjective but not injective?

I know there are injections [e.g. $$\tan(x)$$] and bijections [e.g. the classic $$\tan(\pi (x-\frac{1}{2}))$$] from $$(0, 1)$$ to $$\mathbb{R}$$, but I had been stuck after I tried to construct a function from $$(0, 1)$$ to $$\mathbb{R}$$ that is only surjective but not injective. So I wonder if there is any example. Thanks in advance!

If $$f$$ is a bijection $$(0,1)\to\mathbb R$$, then consider $$g(x) = \begin{cases}f(2x) & x\in(0,1/2), \\ 0& x\in [1/2,1).\end{cases}$$

If you already have a bijection $$f: (0, 1) \to \Bbb R$$ (such as $$f(x) = \tan(\pi (x-\frac{1}{2}))$$) then you can compose it with any function $$g: \Bbb R \to \Bbb R$$ which is surjective but not injective. Then $$g \circ f: (0, 1) \to \Bbb R$$ has the desired properties.

A possible choice for $$g$$ is $$g(x) = x^3 - x$$, or any other odd-degree polynomial which is not strictly increasing.

Sure, it's not mysticism. With infinite sets of equal cardinality you can always tweak things.

Let $$\phi$$ be a bijection from $$(0,1)\to \mathbb R$$. You give the classic of $$\phi(x) = \tan(\pi(x-\frac 12))$$. That only will do nicely.

Then $$\phi': (0,\frac 12) \to \mathbb R$$ via $$\phi'(x) = \phi(2x)$$ is a bijection.

And $$\overline{\phi}:(\frac 12,1)\to \mathbb R$$ via $$\overline{\phi}(x) =\phi'(x-\frac 12) = \phi(2(x-\frac 12))$$ is a bijection.

And $$\gamma:(0,1)\to \mathbb R$$ via $$\begin{cases} \phi'(x) & x < \frac 12\\ \overline{\phi}(x) & x > \frac 12\\e^{\sqrt{27} + \pi} - \frac 2{517}& x = \frac 12\end{cases}$$.

(and if \$w =

$$\gamma$$ is surjective because for any $$w \in\mathbb R$$ then we an $$x=\phi^{-1}(w)\in (0,1)$$ so that $$\phi(x) = w$$. For example if $$\phi: x\mapsto \tan(\pi(x-\frac 12))$$ then $$x=\phi^{-1}(w) = \frac {\arctan (w)}{\pi} + \frac 12$$ will be that $$\phi(x) = w$$.

And so $$\gamma (\frac 12 \phi^{-1}(w))=\phi'(\frac 12 \phi^{-1}(w))=\phi(2\cdot \frac 12 \phi^{-1}(w))=\phi(\phi^{-1}(w)) = w$$ and $$\gamma(\frac 12 \phi^{-1}(w) +\frac 12) = \overline{phi}(\frac 12 \phi^{-1}(w) +\frac 12) = \phi(2(\frac 12 \phi^{-1}(w) +\frac 12-\frac 12))=\phi(\phi^{-1}(w)) = w$$.

So $$\gamma$$ is surejective. But $$\frac 12 \phi^{-1}(w)\ne \frac 12 \phi^{-1}(w) +\frac 12$$ so $$\gamma$$ is not injective.

(and if $$w = e^{\sqrt{27} + \pi} - \frac 2{517}$$ then we have $$\frac 12, \frac 12(\frac {\arctan ( e^{\sqrt{27} + \pi} - \frac 2{517})}{\pi} + \frac 12),$$ and $$12(\frac {\arctan ( e^{\sqrt{27} + \pi} - \frac 2{517})}{\pi} + \frac 12)$$ all mapping to $$w$$.)