Continuous function from $(0,1)$ onto $[0,1]$ 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.
 A: Your example for A is fine.
For B "for a continuous function on an interval, being one-to-one is equivalent to being increasing throughout or decreasing throughout" so that no such function exists.
(see (2.5)-(2.6) of this paper)
A: 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.
A: A slightly simpler solution, perhaps, for the first part would be as follows:
Consider the function from $\left[\frac14,\frac34\right]$ to $[0,1]$ defined by $x\mapsto 2\left(x-\frac14\right)$. This is certainly a continuous function, and it is certainly onto $[0,1]$.
Define now:
$$f(x)=\begin{cases}0 & x<\frac14\\ 2\left(x-\tfrac14\right) & x\in\left[\tfrac14,\tfrac34\right]\\ 1 & x>\frac34\end{cases}$$
Of course this is not an injective function, but it is continuous and onto, as required.
A: Answer for (B): Continuous image of a compact set is compact. [see1]
Suppose $f: [0,1] \rightarrow (0,1)$ is a bijection and continuous. Then $f([0,1])=(0,1)$, which is compact by above theorem. 
But in $\mathbb{R}$, the closed interval $[0,1]$ is compact; but the open interval $(0,1)$ is not compact. 
Therefore if such $f$ does not exist, then $f^{-1}$ does not exist too, where $f^{-1}: (0,1) \rightarrow [0,1]$. (your question in (B))
A: If such a function exists then (0,1) and [0,1] are homeomorphic. But if you delete 0 from [0,1] you get (0,1] which is connected. Deleting the point x in (0,1) which corresponds to 0 in [0,1] leaves you with (0,x) u (x,1) which is not connected ---> contradiction. 
