real-analysis
While revising, I came across this question(s):
A) Is there a continuous function from $(0,1)$ onto $[0,1]$?
B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$?
(clarification: one-to-one is taken as a synonym for injective)
I figured the answer to A is yes, with $\frac{1}{2}\sin(4\pi x)+\frac{1}{2}$ as an example.
The answer to part B is no, but what is the reason?
Sincere thanks for any help.
Try $f(t) = \sin(1/t)$.
If you want to use your absolute value idea, try making a $V$ shape with the vertex at $(\tfrac12, 0)$, opening upwards to include the points $(0,1)$ and $(1,1)$. (I'll leave the actual function up to you.)
I suggest to start by drawing: draw the bounding box $[0,1)\times (-1,1)$, place your pencil at $(0,0)$, and trace different functions. The following are observations and hints, not logical proofs.
A first idea is that, since you want a continuous function, a monotonous one might be troublesome in mapping a semi-open interval like $[0,1)$ to an open one like $(-1,1)$. You can find one location in $[0,1)$ where $f(x)$ approaches $ 1$ and another location in $[0,1)$ where $f(x)$ approaches $ -1$. If these locations are in some $[\epsilon,1-\epsilon]$ ($\epsilon >0$), continuity might impose that the values $-1$ or $1$ will be reached strictly inside $[0,1)$, which you do not want.
One remaining choice is that values $y=-1$ and $y=1$ are both approached on an open end like $\left[.~ 1\right)$. So you might need a function that oscillates infinitely often close to $x=1$. In other word, it can be wise to open $[0,1)$ to something like $[0,+\infty)$.
To summarize, three main ingredients ingredients could be useful, in the shape of a fellowship of continuous functions that you could compose. Many choices are possible, here is one (borrowing from Tolkien's Ring poem):
in the Land of functions where continuity lies (indeed, continuous functions tend to cast continuous shadows to connected intervals).
The first ingredient $f_1(x)$ is easy with functions you know, defined on some $[a,b[$ with a singularity at $b$. It is easy to remap $[a,b[$ to $[0,1[$, so let us stick to that interval $[0,1[$. Examples: $\frac{1}{1-x}$, $-\log (1-x)$, $\tan (\pi x/2)$, and many more.
If you want more flexibility, you can start with any function $f_0$ that maps $[0,1[$ to $[0,1[$: $x^p$ with $p>0$, $\sin(\pi x/2)$, $\log(1+x(e-1))$.
For the second ingredient $f_2$ on $[0,+\infty)$, the sine is a nice pick, and a lot of combinations of sines and cosines, like a chirping sound. But you have a lot of fancy alternatives. And you can easily plug in a function $f_\phi$ that maps $[0,+\infty)$ to $[0,+\infty)$ (for instance $\exp(x)-1$, $x^p$).
The choice of $f_2$ is possibly the most sensitive, since you will need to strictly bound it afterward inside $ (−1,1)$, therefore you will need a third ingredient: a function $f_3$ that compensates (as a product for instance) the envelope of $f_2$ so that the result does not exceed $1$ in absolute value. So for the sine, $x^p$ or $\frac{\exp{x}-1}{e-1}$ will do the job. A function whose magnitude is strictly less than $1$, and tends to $1$ as $x\to 1$ is likely to work.
Finally, a function can be obtained by composing $f(x)= f_3(x)\times f_2\left( f_\phi\left( f_1\left( f_0\left( x\right)\right)\right)\right)$.
In your case, you have for instance $f_0(x)=x$, $f_1(x)=\frac{1}{1-x}$, $f_\phi(x)=x$, $f_2(x)=\sin(x)$, $f_3(x)=x^2$.
While not fully generic, you can cook of lot of recipes with those ingredients. For instance (see image below):$$f(x)=\sin(\pi x^7/2)\sin \left( \tan \left(\pi \sqrt{x}/2\right)\right)\,.$$
Best Answer
There isn't such a function. If $f$ is such a function, then $f$ is monotonic, and $f^{-1}$ too. But a monotonic function defined on an interval is continuous iff its range is an interval. So we have that $f$ is an homeomorphism from $(0,1)$ to $[0,1]$, which is impossible since one is compact and the other is non compact.