F is an open mapping implies f inverse cont.

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SUMMARY

The discussion centers on the proof that if a continuous bijection \( f: X \to Y \) is an open mapping, then its inverse \( f^{-1} \) is continuous. The key argument involves demonstrating that for any open set \( V \) in \( Y \), there exists an open set \( W \) in \( X \) such that \( f[W] \subseteq V \). The participants emphasize the importance of articulating their understanding to clarify their thoughts and solidify their grasp of the proof.

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  • Understanding of continuous functions in topology
  • Familiarity with bijections and their properties
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  • Basic proof techniques in mathematical analysis
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Unassuming
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We have a continuous bijection f:X-->Y.

Prove that if f is open, then f inverse is continuous.


I can't figure it out.
"Proof". For V open in Y, there exists W open in X such that [tex]f[W] \subseteq V[/tex]. Where does the f is open definition apply?
 
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Ah! I got it. Posting always helps the blood get flowing. Trying to explain what I know to others.
 

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