Conditions for being an open map - check my proof

Question

We define an open map as follows:

Let be $f:M\to \mathbb{R}^n$, where $M \subseteq \mathbb{R}^n$. If $f(Q)$ is open for every open set $Q \subseteq M$ then we call $f$ an open map.

Let's take an open set $Q$ and be $y\in f(Q)$ then there exists a $q \in Q$ with $f(q)=y$. As $Q$ is open there exists a neighbourhood $N$ of $q$ with $N\subseteq Q$. We also know that $f(N) \subseteq f(Q)$ and that $f(q) \in f(N)$. So we found a neighbourhoof of $y$ which lies in $f(Q)$. Hence, $f(Q)$ is open.

This means that all functions are open maps. Where is my mistake?!

Answer 1

A constant function will not be an open map since points are not-open in Euclidean spaces. Your mistake is assuming that $f(N)$ is also open: you are making use of what you are trying to prove. For instance, note how your proof breaks down if $f$ were constant on $N$.

Answer 2

The set $f(N)$ need not be open(unless $f$ is an open map).${}{}{}{}$