f is continuous on $(X,\tau)$ iff pre-image of set’s interior is subset of pre-interior of set

Question

$f: (X,\tau_1) \to (Y,\tau_2)$ Show $f$ is continuous on $(X,\tau_1) $ if and only if $\forall B \subseteq Y$ , ${f^-1}(interior(B)) \subseteq interior(f^{-1}(B))$

I don’t have any problem in first implication (let f is continuous in X then pre-image of interior of B is a subset of interior of pre-image of B) but I am not sure about converse. Which open set should I use? I have tried $interior(B)$ but it seems like wrong to me. Could someone show this implication clearly for me?

Answer 1

Proof by contrapositive.

If $f$ is not continuous, there exists some open set $B \in \tau_2$ such that $f^{-1}(B) \notin \tau_1$, i.e. its preimage is not open.

Then, $\text{int}(f^{-1}(B)) \subsetneq f^{-1}(B) = f^{-1}(\text{int}(B))$, so $f^{-1}(\text{int}(B)) \not\subseteq \text{int}(f^{-1}(B))$.

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Direct proof.

Let $B \in \tau_2$ be open. Then $f^{-1}(B) = f^{-1}(\text{int}(B)) \subseteq \text{int}(f^{-1}(B))$ by assumption. Since $f^{-1}(B)$ is contained in its own interior, it must be open.