Do continuous injections preserve open sets?
I'm pretty sure that's true in euclidean space.
If we let the singleton sets of integers generate the topology of the domain, and then identity map it to the real with standard topology, is that a counterexample?
If they don't, what combination of injective, surjective, continuous, and inverse continuous is the minimum to be an open map?
Edit: Either I was really tired and distracted on the bus when I typed this into my phone and somehow forgot to say continuous (entirely possible), or whoever put "from $\mathbb{R}^m$ to $\mathbb{R}^n$" in my title deleted it. Should I start a new one?
Best Answer
Answer to old question before a significant edit was made:
No they do not. It's not true in euclidean space.
The function $$f(x)=\begin{cases}x& x\leq 0\\ x+1 & x>0\end{cases}$$ is an injection, however, $f((-1,1)) = (-1, 0]\cup (1, 2)$ which is not open.