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Proof

(related to Proposition: Bijective Open Functions)

By hypothesis, $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ are topological spaces and $f:X\to Y$ is a bijective function.

"$\Rightarrow$"

• Assume, $f$ is an open function.
• Then, the image $f[A]$ of each open set $A\subset X$ is open in $Y.$
• Since $f$ is bijective, the inverse image equals $A$, formally $f^{-1}[f[A]]=A.$
• It follows that for every open set $f[A]$ in $Y$ the inverse image $f^{-1}[f[A]]$ is open $X,$ since $A$ is open in $X.$
• By definition, the inverse function $f^{-1}$ is continuous.

"$\Leftarrow$"

• Assume, the inverse function $f^{-1}$ is continuous.
• Then for every open set $B\in Y$ the inverse image $f^{-1}[B]$ is open in $X.$
• Since $B$ is bijective, $A:=f^{-1}[B]$ is the only subset $A\subseteq X$ for which $f(A)=B.$
• Thus, $A$ is open in $X.$
• Therefore, for every open subset $A\subseteq X$, the image $f[A]$ is open in $Y.$
• It follows, $f$ is an open function.
  ∎

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References

Bibliography

1. Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970
2. Jänich, Klaus: "Topologie", Springer, 2001, 7th Edition