Let $\Omega \subset \mathbb R^N$ and $\varphi:\Omega \to \mathbb R^n$ an injective and continuous function. Brouwer proved that $\varphi$ is an open application and that $\varphi$ is a homeomorphism. Then the author of the book mentioned as a remark : "Some author assume the strongest hypothesis that $\varphi$ is already an homeomorphism. Even with this assumption, it's hard to prove that $\varphi(\Omega )$ is an open set in $\mathbb R^N$ when $\Omega $ is open"
Now there is things I always thought and that looks wrong.
Q1) First of all, if $\varphi$ is open, then the fact that $$\varphi|^{\varphi(\Omega )}:\Omega \to \varphi(\Omega ),$$ is obviously an homemorphism, right ? Because it's bijective, continuous and open. This is correct ?
Q2) For the part : "Even with this assumption, it's hard to prove that $\varphi(\Omega )$ is an open set in $\mathbb R^N$ when $\Omega $ is open", I don't understand ! If $\varphi :\Omega \to \varphi(\Omega )$ is an homemorphism, is $\varphi(\Omega )$ open ? (I think it's almost definition of an homeomorphism). Indeed, if $\Omega $ is an homeomorphism, then $\varphi$ is open, and thus $\varphi(\Omega )$ is also open. How can it be harder ?
For Q1) yes it's correct.
For $Q2)$ you have that $\varphi(\Omega )$ is open for the topology induced by $\mathbb R^n$, but this doesn't mean that $\varphi(\Omega )$ is open in $\mathbb R^n$. For example, if $\mathcal T$ denote the usual topology on $\mathbb R$, then $[0,1]$ is open for the topology $$\mathcal T_{[0,1]}:=\{U\cap [0,1]\mid U\in \mathcal T\},$$ but it's not open in $\mathbb R$. So saying $\varphi:\Omega \to \varphi(\Omega )$ is a open mean that if $U$ is open in $\Omega $, then $\varphi(U)= \varphi(\Omega )\cap V$ for an open $V$ in $\mathbb R^n$, but it doesn't mean that $\varphi(U)$ is open in $\mathbb R^n$.