$ f $ is open if and only if $f^{-1}(\mathrm{Int}(B))\supset \mathrm{Int}(f^{-1}(B))$

Question

I'm working on the following problem:

$f$ is open if and only if $f^{-1}(\mathrm{Int}(B))\supset \mathrm{Int}(f^{-1}(B))$ for all $B$.

For $\Longrightarrow$: Suppose $f$ is open, let $x\in \mathrm{Int}(f^{-1}(B))$, let $O$ be an open set such that $x\in O\subset f^{-1}(B)$. Then $f(O)$ is an open set contained in $B$, hence $f(x)\in \mathrm{Int}(B)$, $x\in f^{-1}(\mathrm{Int}(B))$.

I'm stack on $\Longleftarrow$, any hints are welcome.

Answer 1

Let $O$ be an open set; you want to prove that $f(O)$ is open. Suppose otherwise. Then there is some $x\in O$ such that $f(x)\notin f(O)^\circ$. In other words, $x\notin f^{-1}\bigl(f(O)^\circ\bigr)$. But $x\in O\subset f^{-1}\bigl(f(O)\bigr)$. Since $O$ is open $x\in\bigl(f^{-1}\bigl(f(O)\bigr)\bigr)^\circ$. So, taking $B=f(O)$ this proves that $f^{-1}(B^\circ)\not\supset\bigl(f^{-1}(B)\bigr)^\circ$.