A simple counter example for Open Mapping Theorem when Linearity Assumption is Dropped I was studying on Follan's Real Analysis book and in Chapter 5.3, I saw the Open Mapping Theorem which stated (with some adjustment) as follows:

Let $X,Y$ be Banach spaces and $T:X\to Y$ be linear and bounded operator. If $T$ is surjective, then $T$ maps open sets to open set.

Then, based on the expression above, I expect if $T$ is nonlinear then the conclusion must fail. So I consider an example as follows: Let $X=Y=\mathbb{R}$ be (simplest?) Banach spaces and then take a mapping $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x^2$. Then consider open interval $U:=(-1,1)$ then $f(U) = [0,1)$ which is not open. Is my thinking correct? 

Consider $f: \mathbb{R} \to \mathbb{R}$ by $f(x)=x^3 -3x^2$.  This is bounded (as it's continuous) and surjective.  However, the open set $(0,3)$ in the domain maps to $[-4,0)$, which isn't open.  
(Aweygan gets the assist.) 
