Restriction of open map on topological space to its subspace is open map

Question

suppose $f:X\to Y$ is open surjection .suppose $S\subset X$ and $f^{-1}[f[S]]=S.$

show that mapping $f|S:S\to f[S]$ is open map. where relative topologies are put on $S$ and $f[S]$.

give an example to show that conclusin is not necessarily true is $S$ is not an inverse set.

what i tried is that let $O$ is open set in $S.$ then $O=S\ \cap O_1$.where $O_1$ is open in $X$.i know $f(O_1)$ is open .but how i connect it to set $f(O)$.

$f(O)=f(S\ \cap O_1) \subset f(S)\ \cap f(O_1)$. and i think other way around is not true. then how to show $f(O)$ is open. any hint ??

Answer 1

Actually $f(S\cap O_{1})=f(S)\cap f(O_{1})$: For $y\in f(S)\cap f(O_{1})$, say, $y=f(u)\in f(S)$, $y=f(v)\in f(O_{1})$, $u\in S$, $v\in O_{1}$, then $v\in f^{-1}(f(S))=S$, so $v\in S$, so $v\in S\cap O_{1}$, so $y=f(v)\in f(S\cap O_{1})$.