Existence of Continuous bijective function I was looking for examples of bijective continuous functions infollowing cases


*

*Does there exist Bijective continuous function $(0,1) \rightarrow \mathbb{R}$?


Yes, $f(x)=\frac{2x-1}{x-x^2}$


*Does there exist Bijective continuous function from $\mathbb{R} \rightarrow  (0,1)$ ?


Yes, as $(0,1) and \mathbb{R}$ are homeomorphic so inverse of $f(x)$ defined above will be example such a function.


*Does there exist a bijective continuous function from $[a,b] \rightarrow \mathbb{R}$?


I think No, as continuous image of compact set is not compact.


*Does there exist a bijective continuous function from $\mathbb{R} \rightarrow  [a,b])$ ?


For this clearly it can be seen 
a) continuous image of connected set is connected.
b) Inverse image of closed set under continuous map is closed.
So there exist such a function but i can not construct it.


*Does  Does there exist a bijective continuous function from $ (a,b)  \rightarrow  [a,b])$?


No. as inverse image of closed set is not closed so there can not exist any continuous bijective function.
But as $(a,b)$ is  homeomorphic  to $\mathbb{R}$So if we construct a function $\mathbb{R \rightarrow}  [a,b]$ will that function work as an example of a function $ (a,b)  \rightarrow  [a,b])$?


*Does  Does there exist a bijective continuous function from $ [a,b] \rightarrow  (a,b))$?


No. as inverse image of open set is not open so there can not exist any continuous bijective function.
Please correct me if I am wrong anywhere .And give examples of $(4)$ if any exists. Thanx very much for your time
 A: 

*No. Suppose there exists such a function $f$. Say $f(x_0) = a$, then $f(x) > a$ for both $x<x_0$ and $x> x_0$, in contradiction with $f$ being a bijection.


5 and 6. Your arguments are invalid. Given continuous function $f:X\to Y$, the pre-image of an open(closed) set in $Y$ under $f$ is open(closed) in $X$. Here $(a,b)$ is indeed closed in $(a,b)$, and $[a,b]$ likewise. That no continuous function maps $[a,b]$ onto $(a,b)$ can be easily seen by noticing that $[a,b]$ is compact while $(a,b)$ isn't. That no continuous bijection maps $(a,b)$ to $[a,b]$ can be reasoned similarly as in 4.
Also be aware that in general a continuous bijection doesn't have to be a homeomorphism, so you probably want to refine your arguments in 2. That said, we do have invariance of domain theorem in Euclidean spaces.
A: Your proposed $f$ for (1) is not a bijection. One that is:
$f(x)=\tan(\pi x-\frac{\pi}{2})$
Crucially, any such $f$ must be monotone (in-fact it cannot have derivative 0 over any interval, only at single points) (why?)
For (2), just invert (the corrected) function for #1.
(3) is no because $[a,b]$ is compact, but $\mathbb{R}$ is not (cannot satisfy surjectivity and continuity at the same time for reason you gave).
(4) is also no, but the reasoning is slightly different (see @QiyuWen's answer).
(5) You're reasoning is wrong, it's for the same reason as in (4) (this has to do w/ monotonicity again).
(6) Appeal to compactness again ($(a,b)$ is not compact; why?).
