(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.

- 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.

- 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.∎

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