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F is an open mapping implies f inverse cont.

  1. Nov 21, 2008 #1
    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?
  2. jcsd
  3. Nov 21, 2008 #2
    Ah! I got it. Posting always helps the blood get flowing. Trying to explain what I know to others.
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