Non-existence of bijective, continuous function from $(0,1)$ to $[0,1]$ The problem : Give example of a continuous, onto function from $(0,1)$ to $[0,1]$. Is it possible for such a function to be one-one?
My partial solution : For the $1^{st}$ part of the question, I came up with this example -
$f : (0,1) \to [0,1]$ given by $f(x)=\sin\big({2\pi x}\big)$
This is continuous and onto, but not one-one.
What I'm asking : For the $2^{nd}$ part of the question, I feel that it should be provable that there cannot exist a continuous, onto, one-one function from $(0,1)$ to $[0,1]$ (If not, we need a counter-example). Any help regarding this proof (or what would really surprise me, a counter example)?
Thanks in advance.
 A: Bijective continuous map implies the function is monotone, now $f(x)=0$ for some $x\in (0,1)$.
We know there exists a real number $y\in (0,x)$ and a real number  $z\in (x,1)$.
Now since $f(x)=0$ and the function is bijective then both $f(y)$ and $f(z)$ are greater than $0$ so the function was decreasing somewhere in $(y,x)$ and increasing somewhere in $(x,z)$, contradicting the monotonicity of $f$. 
A: Suppose $f$ is bijective, then $f$ must be strictly increasing or decreasing function. W.l.g. take $f$ to be increasing. Then $\exists x\in (0,1)$ such that $f(x)=1$. Now take $x<y<1$, since $f$ is increasing, $f(y)>1$, which leads to contradiction.
A: Assume there exists $x \in (0,1)$ such that $f(x) = 0$. Then, by openness of $(0,1)$, for some $\epsilon > 0$, the interval $(x-\epsilon, x+\epsilon)$ is in the interval $(0,1)$. 
If now $f$ is one-to-one and onto, either $f(x-\epsilon) < f(x)$ or $f(x+\epsilon) < f(x)$. This contradicts the fact that we want $f$ to map to the closed interval $[0,1]$.
A: If $f: A\rightarrow B$ is a continuous map, then $f$ pulls closed sets in $B$ to closed sets in $A$. Given, $f$ is onto and $[0, 1]$ is closed. So its pre-image should be closed. But its pre-image $(0,1)$ is not cloed in $\mathbb{R}$. A contradiction. So we can't have such $f$
