Do continuous injections preserve open sets?

Question

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?

Answer 1

Another example : take a set $X$ with at least two elements. Consider $X_1$ to be $X$ equiped with the discrete topology and $X_2$ to be $X$ equiped with the trivial topology. Then, the identity from $X_1$ to $X_2$ is clearly a continuous injection (it is even a bijection) but it clearly does not preserve open sets.

I don't think there is a "minimum" requirement on a map to be open, it depends mostly on the topologies. (But homeomorphisms are always open).

Answer 2

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.

Answer 3

Whether $f:X\to Y$, and in particular an injection, is an open map depends on the topologies, e.g., if $Y$ has the discrete topology, then any $f$ is an open map. You can check that if $f$ is injective onto an open subset of $Y$ and $f^{-1}$ is continuous, then $f$ is an open map.

Answer 4

It is not true over multiple Euclidean spaces.

Just consider the natural projection $f: \mathbb R \to \mathbb R^2$ where $f(x) = \left[\begin{matrix}x \\ 0\end{matrix}\right]$. $f(S)$ is never open for $\emptyset \ne S \subset \mathbb R$, because for any $x \in S$, $f(x) + \left[\begin{matrix}0 \\ r\end{matrix}\right]$ is not in the open ball $B(f(x), 2r)$. Illustratively, $f((0, 1))=(0,1) \times \{0\}$ which is obviously not open (nor closed).

Answer 5

Isn't the case $\mathbb R^n \to \mathbb R^n$ a result of L.E.J. Brouwer? In any case: $\mathbb R^1 \to \mathbb R^1$ is easy from calculus facts, while $\mathbb R^2 \to \mathbb R^2$ is harder, but can be proved from the Jordan curve theorem.

added
I remembered the name, "Invariance of domain"
See https://en.wikipedia.org/wiki/Invariance_of_domain