Does $f^{-1}(B^\circ)\subset (f^{-1}(B))^\circ$ imply continuity? Let $(X,\tau),(Y,\eta)$ be topological spaces and  $f:X\to Y$ a function. Prove that  $f$ is continuous if and only if $f^{-1}(B^\circ)\subset (f^{-1}(B))^\circ,\forall B\subset Y$
$\implies$
Let $B\subset Y.$ As $f$ is continuous, $f^{-1}(B^\circ)=(f^{-1}(B^\circ))^\circ\subset (f^{-1}(B))^\circ$
$\Longleftarrow$
What can I do to prove this side?
Is the implication $\implies$ correct?
 A: For an open set $G$ of $Y$, $f^{-1}(G)=f^{-1}(G^{\circ})\subseteq(f^{-1}(G))^{\circ}\subseteq f^{-1}(G)$, so $f^{-1}(G)=(f^{-1}(G))^{\circ}$.
A: Your first implication is already too terse, I think. 
Suppose that $f$ is continuous. Let $B \subseteq Y$.
Then $B^\circ$ is open and $B^\circ \subseteq B$. So $f^{-1}[B^\circ]$ is open as $f$ is continuous. Also $f^{-1}[B^\circ] \subseteq f^{-1}[B]$. So as $(f^{-1}[B])^\circ$ is the union of all open subsets of $f^{-1}[B]$ (the definition of interior) we thus have $f^{-1}[B^\circ] \subseteq (f^{-1}[B])^\circ$, as required. 
A more "axiomatic" way (using the axioms for an interior operator) would be :
$$f^{-1}[B^\circ] = f^{-1}[B^\circ]^\circ \subseteq f^{-1}[B]^\circ$$
where the first equality just says that $f^{-1}[B^\circ]$ is open and the inclusion follows as interiors respect inclusions).
This shows $\Rightarrow$. 
For $\Leftarrow$: Suppose that for all $B \subseteq Y$, we have $f^{-1}[B^\circ] \subseteq f^{-1}[B]^\circ$. 
We want to show $f$ is continuous so take $O$ to be an arbitrary open subset of $Y$, and we need to show $f^{-1}[O]$ is open in $X$.
Then $O^\circ = O$ and so
$$f^{-1}[O]^\circ \subseteq f^{-1}[O] = f^{-1}[O^\circ] \subseteq f^{-1}[O]^\circ \subseteq f^{-1}[O]$$
where the first and last inclusions follow from $A^\circ \subseteq A$ for all $A$, we then use $O =O^\circ$ and then the identity on the right hand side in our assumption, for $B=O$. All inclusions together show $f^{-1}[O]^\circ  =f^{-1}[O]$ so $f^{-1}[O]$ is open.
