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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?

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    $\begingroup$ A bijective map is open iff its inverse is continuous. $\endgroup$ – Kavi Rama Murthy Apr 4 at 7:21
  • $\begingroup$ @KaviRamaMurthy Thanks! If we remove surjectivity, is there another condition we could add to give us an open map? And, if we add bicontinuous, can we remove anything? $\endgroup$ – user3146 Apr 4 at 16:49
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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.

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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.

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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).

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