[proofplan]
We prove the cyclic chain of implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. The implication $(1) \Rightarrow (2)$ uses the [Neighbourhood Characterisation of Closure](/theorems/1005): if $x \in \overline{A}$, every open neighbourhood of $f(x)$ pulls back to a neighbourhood of $x$ (by continuity), which must meet $A$. The implication $(2) \Rightarrow (3)$ applies the closure condition to $A = f^{-1}(F)$. The implication $(3) \Rightarrow (1)$ recovers the open-preimage definition of continuity by complementation.
[/proofplan]
[step:Show $(1) \Rightarrow (2)$: continuity implies $f(\overline{A}) \subset \overline{f(A)}$]
Let $f$ be continuous and let $A \subset X$. Take $x \in \overline{A}$. We show $f(x) \in \overline{f(A)}$ using the [Neighbourhood Characterisation of Closure](/theorems/1005).
Let $V \in \tau_Y$ with $f(x) \in V$. Since $f$ is continuous, $f^{-1}(V) \in \tau_X$ and $x \in f^{-1}(V)$. Since $x \in \overline{A}$, the [Neighbourhood Characterisation of Closure](/theorems/1005) gives $f^{-1}(V) \cap A \neq \varnothing$. Choose $a \in f^{-1}(V) \cap A$. Then $f(a) \in V$ and $f(a) \in f(A)$, so $V \cap f(A) \neq \varnothing$. Since $V$ was arbitrary, $f(x) \in \overline{f(A)}$.
[guided]
Let $f$ be continuous and fix $A \subset X$. We must show that $f(\overline{A}) \subset \overline{f(A)}$, i.e., for every $x \in \overline{A}$, the image $f(x)$ belongs to $\overline{f(A)}$.
Fix $x \in \overline{A}$. By the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $(Y, \tau_Y)$, to show $f(x) \in \overline{f(A)}$ it suffices to show that every open neighbourhood $V$ of $f(x)$ in $Y$ satisfies $V \cap f(A) \neq \varnothing$.
Let $V \in \tau_Y$ with $f(x) \in V$. By continuity of $f$, the preimage $f^{-1}(V)$ is open in $X$, and $x \in f^{-1}(V)$ since $f(x) \in V$. By the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $(X, \tau_X)$, since $x \in \overline{A}$, we have $f^{-1}(V) \cap A \neq \varnothing$.
Pick $a \in f^{-1}(V) \cap A$. Then $f(a) \in V$ (because $a \in f^{-1}(V)$) and $f(a) \in f(A)$ (because $a \in A$). Hence $f(a) \in V \cap f(A)$, so $V \cap f(A) \neq \varnothing$.
[/guided]
[/step]
[step:Show $(2) \Rightarrow (3)$: the closure condition implies preimages of closed sets are closed]
Suppose $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$. Let $F \subset Y$ be closed. Set $A := f^{-1}(F)$. Applying the hypothesis:
\begin{align*}
f(\overline{f^{-1}(F)}) \subset \overline{f(f^{-1}(F))} \subset \overline{F} = F,
\end{align*}
where the second inclusion uses $f(f^{-1}(F)) \subset F$ and monotonicity of closure, and the equality uses the fact that $F$ is closed. Therefore $\overline{f^{-1}(F)} \subset f^{-1}(F)$. Combined with extensivity $f^{-1}(F) \subset \overline{f^{-1}(F)}$, we get $\overline{f^{-1}(F)} = f^{-1}(F)$, so $f^{-1}(F)$ is closed.
[guided]
Suppose condition (2) holds: $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$. Let $F \subset Y$ be closed; we show $f^{-1}(F)$ is closed in $X$.
Set $A := f^{-1}(F)$ and apply condition (2):
\begin{align*}
f(\overline{f^{-1}(F)}) \subset \overline{f(f^{-1}(F))}.
\end{align*}
The set $f(f^{-1}(F))$ is contained in $F$ (for any map $f$ and any set $F$, $f(f^{-1}(F)) \subset F$). By monotonicity of closure ($S \subset T$ implies $\overline{S} \subset \overline{T}$), we get $\overline{f(f^{-1}(F))} \subset \overline{F}$. Since $F$ is closed, $\overline{F} = F$. Combining:
\begin{align*}
f(\overline{f^{-1}(F)}) \subset F.
\end{align*}
Taking preimages: if $x \in \overline{f^{-1}(F)}$, then $f(x) \in F$, so $x \in f^{-1}(F)$. This gives $\overline{f^{-1}(F)} \subset f^{-1}(F)$. Extensivity gives the reverse inclusion, so $\overline{f^{-1}(F)} = f^{-1}(F)$, meaning $f^{-1}(F)$ is closed.
[/guided]
[/step]
[step:Show $(3) \Rightarrow (1)$: closed preimages implies continuity]
Suppose $f^{-1}(F)$ is closed in $X$ for every closed $F \subset Y$. Let $V \in \tau_Y$. Then $Y \setminus V$ is closed in $Y$, so $f^{-1}(Y \setminus V) = X \setminus f^{-1}(V)$ is closed in $X$. Therefore $f^{-1}(V)$ is open in $X$. Since $V$ was arbitrary, $f$ is continuous.
[/step]
