# 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 ?

• Such examples must be infinite-dimensional (due to invariance of domain). – Henno Brandsma Mar 30 at 23:25

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 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 $$(0,1]$$.