[proofplan]
We show equivalence of continuity (preimages of open sets are open) with the interior inclusion $f^{-1}(A^\circ) \subset (f^{-1}(A))^\circ$ for all $A \subset Y$. For $(1) \Rightarrow (2)$: continuity makes $f^{-1}(A^\circ)$ open, and since $A^\circ \subset A$ implies $f^{-1}(A^\circ) \subset f^{-1}(A)$, the set $f^{-1}(A^\circ)$ is an open subset of $f^{-1}(A)$, hence contained in the interior. For $(2) \Rightarrow (1)$: given an open set $V \subset Y$, apply the hypothesis with $A = V$ and use $V^\circ = V$ to conclude $f^{-1}(V) \subset (f^{-1}(V))^\circ$, which forces $f^{-1}(V)$ to equal its interior and hence be open.
[/proofplan]
[step:Show that continuity implies $f^{-1}(A^\circ) \subset (f^{-1}(A))^\circ$ for all $A \subset Y$]
Let $f: X \to Y$ be continuous and let $A \subset Y$. The set $A^\circ$ is open in $Y$. Since $f$ is continuous, $f^{-1}(A^\circ)$ is open in $X$. From $A^\circ \subset A$ (property (1) of the [Properties of the Interior Operator](/theorems/1013)), monotonicity of preimages gives $f^{-1}(A^\circ) \subset f^{-1}(A)$. Thus $f^{-1}(A^\circ)$ is an open subset of $f^{-1}(A)$, and by definition of the interior as the largest open subset:
\begin{align*}
f^{-1}(A^\circ) \subset (f^{-1}(A))^\circ.
\end{align*}
[guided]
The argument hinges on two facts: (i) continuous maps pull back open sets to open sets, and (ii) the interior of a set contains every open subset. Applying (i) to the open set $A^\circ \subset Y$ gives $f^{-1}(A^\circ) \in \tau_X$. The inclusion $A^\circ \subset A$ is a set-theoretic fact about the interior (property (1) of the [Properties of the Interior Operator](/theorems/1013)), and the general identity $B \subset C \implies f^{-1}(B) \subset f^{-1}(C)$ applied with $B = A^\circ$, $C = A$ gives $f^{-1}(A^\circ) \subset f^{-1}(A)$. So $f^{-1}(A^\circ)$ is an open set contained in $f^{-1}(A)$. Since $(f^{-1}(A))^\circ$ is by definition the union of all open subsets of $f^{-1}(A)$, it contains $f^{-1}(A^\circ)$.
[/guided]
[/step]
[step:Show that the interior inclusion for all $A \subset Y$ implies continuity]
Assume $f^{-1}(A^\circ) \subset (f^{-1}(A))^\circ$ for every $A \subset Y$. Let $V \in \tau_Y$ be an arbitrary open set in $Y$. We must show $f^{-1}(V) \in \tau_X$.
Since $V$ is open, $V = V^\circ$ by the [Open Set Characterisation](/theorems/1016). Applying the hypothesis with $A = V$:
\begin{align*}
f^{-1}(V) = f^{-1}(V^\circ) \subset (f^{-1}(V))^\circ.
\end{align*}
The reverse inclusion $(f^{-1}(V))^\circ \subset f^{-1}(V)$ holds by property (1) of the [Properties of the Interior Operator](/theorems/1013). Combining:
\begin{align*}
f^{-1}(V) = (f^{-1}(V))^\circ.
\end{align*}
By the [Open Set Characterisation](/theorems/1016), the equality $f^{-1}(V) = (f^{-1}(V))^\circ$ implies that $f^{-1}(V)$ is open. Since $V \in \tau_Y$ was arbitrary, $f$ is continuous.
[guided]
The strategy is to reduce to the [Open Set Characterisation](/theorems/1016): a set equals its interior if and only if it is open. Given an open set $V$ in $Y$, we apply the hypothesis with $A = V$. The left-hand side becomes $f^{-1}(V^\circ) = f^{-1}(V)$ because $V$ is open and therefore $V = V^\circ$. The right-hand side gives $(f^{-1}(V))^\circ$. So we obtain $f^{-1}(V) \subset (f^{-1}(V))^\circ$. But the interior of any set is contained in that set, so $(f^{-1}(V))^\circ \subset f^{-1}(V)$. Together these give equality, and the [Open Set Characterisation](/theorems/1016) then says $f^{-1}(V)$ is open. Since this holds for every open $V \subset Y$, the map $f$ is continuous by the topological definition of continuity.
[/guided]
[/step]
