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My doubts concerning this question appeared in the context of differential geometry, and my knowledge of topology is limited. My attempts at counterexamples all ended up being based on a non open domain, so I believed my conjecture was true.

I came to the conclusion (using the metric topology, the only one I know) it is equivalent to the continuity of the inverse function (and a quick search after confirmed this), but the counterexamples that use the metric topology (like Functions which are Continuous, but not Bicontinuous) all use non open domains.

The question is: Let $f:U \rightarrow V$, with both $U$ and $V$ open sets (of $R^n$, if needed) be a continuous bijection. Is $f(A)$, where $A \subset U$ is an open set, an open set?

RicardoMM
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    Specifically if $U$ and $V$ are open subsets of $\mathbb{R^n}$ then this is true. This is a very non trivial result called invariance of domain, and its proofs require some advanced tools. – Mark Apr 11 '23 at 14:01
  • @Mark I saw it now on wikipedia, thanks. I thought it should be a more trivial result. Please post your comment as answer for me to accept it – RicardoMM Apr 11 '23 at 14:06
  • What do you mean by open domain? All topological spaces are open as subsets of themselves. – Thomas Andrews Apr 11 '23 at 14:27
  • @ThomasAndrews Because I didn't mean to define a function in a topological space, but on an open subset of it. I could define a function on a closed subset of the space, for exemple. – RicardoMM Apr 11 '23 at 15:01
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    But every function on an open subset of a topological space is a function on a topological space. And every topological space is an open subset of some other typological space. You seem to want open subsets of a particular type of topological space, like $\mathbb R^n.$ "Open domain" is meaningless, unless you are talking about subsets of a particular space. – Thomas Andrews Apr 11 '23 at 15:17

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This is true specifically for open subsets of $\mathbb{R^n}$. Even more can be said: if $U\subseteq\mathbb{R^n}$ is open, and $f:U\to\mathbb{R^n}$ is an injective, continuous function, then $f(U)$ is open in $\mathbb{R^n}$ (that is, you don't even need to assume that, it follows), and $f$ is an open map. This nontrivial result is called invariance of domain.

Mark
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In general, this is false. Take $X$ to be any set with at least two elements and let $\mathscr T_1=\mathcal P(X)$ be the discrete topology and $\mathscr T_2=\{X,\emptyset\}$ the trivial topology.

Let $f:(X,\mathscr T_1)\to (X,\mathscr T_2)$ be the identity function. Then $f$ is continuous (every function is), bijective, and given any nontrivial $X_0\subset X$ we have $X_0$ open and $f(X_0)$ not open.

Martin Argerami
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