Let $A: X \rightarrow Y$ a bijective continous map between two Banach spaces X and Y. Then, $A^{-1}$ is also continous?

Question

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 ?

Answer 1

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.

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

3. 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 $(0,1]$.