Let $A: X \rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^{-1}$ is also continous? We know if A is a map continuous, bijective and linear then the answer is yes, $A^{-1}$ is continuous. But, if $A$ is no linear then $A^{-1}$ is also continuous  ?
 A: Let $X$ be an infinite-dimensional Banach space. Then the unit sphere $S\subset X$ is noncompact. Hence, there is a continuous surjective function $h: S\to (0,1]$. For $x\in X-\{0\}$ set $\bar{x}:= x/||x||$. 
Define the self-map
$$
f: X\to X, ~~f(x)=h(\bar{x}) x, ~~\hbox{if}~ x\ne 0; ~~f(0)=0. 
$$ 
This map is continuous and bijective but is not a homeomorphism: Take a sequence $s_n\in S$ such that $\lim_{n\to\infty} h(s_n)=0$. Then the sequence $(f(s_n))$ converges to $0$ while $(s_n)$ does not.  
Edit. 1. If $X, Y$ are finite-dimensional Banach spaces then every continuous bijective map $X\to Y$ is a homeomorphism by Brouwer's invariance of domain theorem. 


*Every infinite dimensional normed vector space has noncompact unit sphere. This was discussed many times at MSE, see for instance here. 

*Every noncompact metric space admits a continuous surjective function to ${\mathbb R}$; this was again discussed many times, see for instance here and here.  Composing with the function $t\mapsto e^{-t^2}$ gives a surjective continuous function to $(0,1]$. 
