Does there exists a onto continuos function $f\colon (0,1) \to[0,1]$ which is one to one . Iam trying to solve this problem , i can visualize graphically it is not possible. iam trying to do the proof by contradiction. if f:(0,1) to [0,1] is a continuos onto function  then i have to prove that f can never be 1-1 . so iam letting that if f is 1-1 then either f is strictly increasing or strictly decreasing function. so case (1) if f is strictly increasing function then there exits some t in (0,1) such that f(t)=0 , so for x lesser than t f(x)k , f(x) &lt;0 which is also not possible .so f can't be 1-1 . My method is right ??? is there is any alternative approach then plz tell
This is merely an elaboration of the solution provided by Júlio Cesar under the original comment (all credit to him), which is in itself "only" making precise your original correct reasoning.  

Assume $f:(0,1) \to [0,1]$ was continuous and bijective. Then we find unique $a := f^{-1}(0)$, $b := f^{-1}(1)$ and may assume wLog that $0 &lt; a &lt; b &lt; 1$. Choose $x \in (0,a)$ and let $y := f(x) \in [0,1]$. By the intermediate value theorem, we find some $c \in (a,b)$ with $f(c) = y = f(x)$. Since $x &lt; a &lt; c$, $f$ is not injective, so we arrive at a contradiction.
$[0,1] $ is closed.

$f $ continuous and one to one implies $f^{-1}([0,1]) $ is closed.

this is not the case for $(0,1) $

Using sequential characterisation of the limit, it is easy to prove the continuity of $f^{-1} $.