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Let $f$ be an open and perfect mapping from $X$ to $Y$. Let $A \subset Y$ and let $y \in \overline{A}$, then whether it is true that $f^{-1}(y) \subset \overline{f^{-1}(A)}$.

I have assumed that $y \in \overline{A} \setminus A$. Let $x = f^{-1}(y)$ i.e. $f(x) = y$. Let $U$ be any open set containing $x$, then I need to show that $U$ intersects with ${f^{-1}(A)}$.

Now $f(x) \in f(U)$ which implies that $f(U) \cap A \neq \emptyset$

i.e. $f^{-1}(f(U) \cap A) \neq \emptyset$

i.e. $f^{-1}(f(U)) \cap f^{-1}(A) \neq \emptyset$.

But this doesn't imply that $U$ intersects with $f^{-1}(A).$

Am I missing some concept to get the required result?

Please clarify on this.

Sumit Mittal
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  • If f is open, then $f^{-1}(\overline{A}) \subset \overline{f^{-1}(A)}$ for each subset $A$ of $Y$. No perfectness needed, not even continuity. – Ulli Dec 12 '24 at 12:00
  • In your above proof, just consider $u \in U$ such that $f(u) \in A$. – Ulli Dec 12 '24 at 12:05
  • Thanks Ulli. I need to ask one more thing whether the inverse image of compact sets is compact under open maps. – Sumit Mittal Dec 13 '24 at 13:24
  • No, of course not: $f: \mathbb N \rightarrow {0}$. – Ulli Dec 14 '24 at 14:13
  • Can you give any example of a bijective mapping?? – Sumit Mittal Dec 15 '24 at 12:51
  • Before you ask again and again, what is your definition of open map? And what is your final question? – Ulli Dec 16 '24 at 22:16
  • Is it true that the inverse image of compact is compact under bijective open maps???? – Sumit Mittal Dec 30 '24 at 07:54
  • Yes, of course: the inverse map of an open, bijective map is continuous. This holds for both possible definitions of openness, which you still haven't specified. – Ulli Dec 31 '24 at 07:05

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