Is there a bijective map from $(0,1)$ to $\mathbb{R}$? I couldn't find a bijective map from $(0,1)$ to $\mathbb{R}$. Is there any example?
 A: Here is a bijection from $(-\pi/2,\pi/2)$ to $\mathbb{R}$:
$$
f(x)=\tan x.
$$
You can play with this function and solve your problem.
A: $g(x)=\frac 1{1+e^x}$ gives a bijection from $\Bbb R$ to $(0,1)$, so take the inverse of this map.
A: A homeomorphism (continuous bijection with a continuous inverse) would be $f:(0,1)\to\Bbb R$ given by $$f(x)=\frac{2x-1}{x-x^2}.$$

Added: Let me provide some explanation of how I came by this answer, rather than simply leave it as an unmotivated (though effective) formula and claim in perpetuity.
Many moons ago, I was assigned the task of demonstrating that the real interval $(-1,1)$ was in bijection with $\Bbb R.$ Prior experience with rational functions had shown me graphs like this:

The above is a graph of a continuous function from most of $\Bbb R$ onto $\Bbb R.$ This doesn't do the trick on its own, since it certainly isn't injective. However, it occurred to me that if we restrict the function to the open interval between the two vertical asymptotes, we get this graph, instead:

This graph is of a continuous, injective (more precisely, increasing) function from a bounded open interval of $\Bbb R$ onto $\Bbb R.$ This showed that rational functions could do the job. Other options occurred to me, certainly (such as trigonometric functions), but of the ideas I had (and given the results I was allowed to use) at the time, the most straightforward approach turned out to be using rational functions.
Now, given the symmetry of the interval $(-1,1)$ (and, arguably, of $\Bbb R$) about $0,$ the natural choice of the unique zero of the desired function was $x=0.$ In other words, I wanted $x$ to be the unique factor of the desired rational function's numerator that could be made equal to $0$ in the interval $(-1,1)$--meaning that for $\beta\in(-1,1)$ with $\beta\ne0,$ I needed to make sure that $x-\beta$ was not a factor of the numerator. For simplicity, I hoped that I could make $x$ the only factor of the numerator that could be made equal to $0$ at all--that is, I hoped that I could have $\alpha x$ as the numerator of my function for some nonzero real $\alpha.$
In order to get the vertical asymptotic behavior I wanted on the given interval--that is, only at the interval's endpoints--I needed to make sure that $x=\pm1$ gave a denominator of $0$--that is, that $x\mp1$ were factors of the denominator--and that for $-1<\beta<1,$ $\beta$ was not a zero of the denominator--that is, that $x-\beta$ wasn't a factor of the denominator. For simplicity, I hoped that I could make $x\mp1$ the only factors of the denominator.
Playing to my hopes, I assumed $\alpha$ to be some arbitrary nonzero real, and considered the family of functions $$g_\alpha(x)=\frac{\alpha x}{(x+1)(x-1)}=\frac{\alpha x}{x^2-1},$$ with domain $(-1,1).$ It was readily seen that all such functions are real-valued and onto $\Bbb R.$
I wanted more, though! (I'm demanding of my functions when I can be. What can I say?) I wanted an increasing function. I determined (through experimentation which suggested proof) that $g_\alpha$ would be increasing if and only if $\alpha<0.$ Again, for convenience, I chose $\alpha=-1,$ which gave me the function that satisfied the desired (and required) properties: $g:(-1,1)\to\Bbb R$ given by $$g(x)=\frac{-x}{x^2-1}=\frac{x}{1-x^2}.$$
Much later, you posted your question, and I realized (again, based on experience) that my earlier result could be adapted to the one you wanted. Playing around a bit with linear interpolation showed that the function $h:(0,1)\to(-1,1)$ given by $h(x)=2x-1$ was a bijection--in fact, an increasing bijection.
It is readily shown (and I had previously seen) that if $X,Y,$ and $Z$ are ordered sets and if we are given increasing maps $X\to Y$ and $Y\to Z,$ then the composition of those maps is an increasing map $X\to Z.$ Also, it is readily shown (and I had seen previously) that if both such maps are continuous and surjective, then so is their composition. Just from these results, my originally posted map was obtained (though named differently): $$\begin{align}(g\circ h)(x) &= g\bigl(h(x)\bigr)\\ &= \frac{h(x)}{1-\left(h(x)\right)^2}\\ &=\frac{2x-1}{1-(2x-1)^2}\\ &=\frac{2x-1}{1-\left(4x^2-4x+1\right)}\\ &= \frac{2x-1}{4x-4x^2}\\ &=\frac{2x-1}{4(x-x^2)}.\end{align}$$
As lhf astutely pointed out shortly thereafter (and as I should have seen immediately), the factor of $4$ in the denominator serves no particular purpose, hence its later removal to yield the function $f$ that I eventually posted.
The remaining claim that I made (that $f$ has a continuous inverse), I leave to you (the reader). If you're curious how I determined this, try to prove it on your own first. If you're stymied (or if you simply want to run your proof attempt by me), let me know. I will do what I can to get you "unstuck."
A: The trigonometric function $\tan x$ is an invertible function from $(-\pi/2,\pi/2)$ to $\mathbb{R}$. Also to find an invertible function from $(0,1)$ to $(-\pi/2,\pi/2)$ find the equation of the straight line joining the points $(0,-\pi/2)$ and $(1,\pi/2)$. Now compose the two functions together. You can likewise find bijections between any two open intervals and any open interval and $\mathbb{R}$.
A: $x \mapsto \ln (- \ln x)$ with the inverse $y \mapsto e^{-e ^ {\ y}}$. It's also a $C ^ \infty$ diffeomorphism.
A: $$(0,1) \overset{\theta \mapsto e^{i 2\pi \theta}}\cong S^1\backslash \{1 \} \overset{e^{i\theta} \mapsto {e^{i(\theta+\pi/2)}}}\cong S^1\backslash \{i \}  \overset{e^{i\theta}\mapsto \frac{cos \theta} {1-\sin\theta}}\cong \mathbb{R} $$The composition of the exponential map, rotation map and stereographic projection is the required bijection.
A: For a less differentiable example, consider the bijection in the following picture,

In symbols, given $x \in (0,1)$ let $n$ be the largest natural number such that $1-\frac{1}{n}<x$, define 
$$y=\frac{x-n}{\frac{1}{n}-\frac{1}{n+1}}$$
to be the renormalized version of $x$ if the interval $(1-\frac{1}{n},1-\frac{1}{n+1}]$ is rescaled and shifted to map to $(0,1)$. Then we have the following bijection:
$$f(x)=\begin{cases}\frac{n-1}{2}+y,& n \text{ odd} \\
-\frac{n-2}{2}-y,& n \text{ even}\end{cases}$$
A: Here is a nice one ${}{}{}{}{}{}{}{}{}$, can you find the equation? 
A: Yes. let $f(x)=\tan((x-1/2)\pi)$. the domain is $(0,1)$ and range is $\mathbb{R}$
A: Yes, see above answers. There are even bijective maps between $(0,1)$ and $\mathbb{R}^n$. To see this, note that a bijection $\phi$ between $(0,1)$ and $(0,1)^2$ can be made in this way: Let $x= 0.b_1b_2\ldots$, with $b_j$ being the digits in a decimal expansion. Define 
$$\phi(x) = (0.b_1b_3b_5\ldots,0.b_2b_4b_6\ldots),$$
i.e., extract even and odd digits. For $\phi^{-1}(x_1,x_2)$, let $x_1 = 0.a_1a_2a_3\ldots$, and $x_2=b_1b_2b_3\ldots$. Then,
$$ \phi^{-1}(x,y) = 0.a_1b_1a_2b_2\cdots$$
Some care has to be taken with identification between digital expansions like $0.199999\cdots$ and $0.20000\cdots$, but that is an exercise.
Having the bijection between $(0,1)$ and $(0,1)^2$, we can apply one of the other answers to create a bijection with $\mathbb{R}^2$.
The argument easily generalizes to $\mathbb{R}^n$.
A: Here is a general bijective function from any interval $(a,b)$ to $\mathbb{R}$:
$$f:(a,b)\to\mathbb{R}, \text{ defined as }f(x)=\tan\left[\frac{\pi}{b-a}\left(x-\frac{b+a}{2}\right)\right]$$
A: The phase shift and periodic reduce tangent function: $tan(x\pi+\frac{\pi}{2})$ maps $(0,1)$ interval to $\mathbb{R}$.
Because it is continuous, monotone and it's range is $(-\infty,+\infty)$.
