First we establish some details. Let $H \subseteq X_1$. Let $\mathbb K_1$ be defined as: :$\mathbb K_1 := \set {K \subseteq X_1: H \subseteq K, K \text { closed} }$ That is, let $\mathbb K_1$ be the set of all closed sets of $T_1$ which contain $H$. Similarly, let $\mathbb K_2$ be defined as: :$\mathbb K_2 := \set {K \subseteq X_2: f \sqbrk H \subseteq K, K \text{ closed} }$ That is, let $\mathbb K_2$ be the set of all closed sets of $T_2$ which contain $f \sqbrk H$. From the definition of closure, we have that: :$\displaystyle H^- = \bigcap \mathbb K_1$ That is, the closure of $H$ is the intersection of all the closed sets of $T_1$ which contain $H$. Similarly: :$\displaystyle \paren {f \sqbrk H}^- = \bigcap \mathbb K_2$ That is, the closure of $f \sqbrk H$ is the intersection of all the closed sets of $T_2$ which contain $f \sqbrk H$. We have: {{begin-eqn}} {{eqn | l = f \sqbrk {H^-}       | r = f \sqbrk {\bigcap \mathbb K_1}       | c = }} {{eqn | o = \subseteq       | r = \bigcap_{K \mathop \in \mathbb K_1} f \sqbrk K       | c = Image of Intersection under Mapping }} {{end-eqn}} From Image of Subset under Relation is Subset of Image: Corollary 2: :$H \subseteq K \implies f \sqbrk H \subseteq f \sqbrk K$ === Necessary Condition === Suppose $f$ is continuous. From the above we have that : :$\displaystyle \paren {f \sqbrk H}^- := \bigcap \mathbb K_2$ As $f$ is continuous, then: :$\forall K \in \mathbb K_2: f^{-1} \sqbrk K$ is closed in $T_1$ But as $f \sqbrk H \subseteq K$, it follows from Image of Subset under Relation is Subset of Image: Corollary 3 that: :$H \subseteq f^{-1} \sqbrk K$ So: :$\mathbb K_3 := \set {f^{-1} \sqbrk K: K \text { closed in } T_2, H \subseteq f^{-1} \sqbrk K}$ consists entirely of closed sets in $T_1$ which are supersets of $H$. That is, $\mathbb K_3 \subseteq \mathbb K_1$. So: :$\displaystyle \bigcap \mathbb K_1 \subseteq \bigcap \mathbb K_3$ and so: :$\displaystyle f \sqbrk {\bigcap \mathbb K_1} \subseteq f \sqbrk {\bigcap \mathbb K_3}$ But from Image of Intersection under Mapping: :$\displaystyle f \sqbrk {\bigcap \mathbb K_3} \subseteq \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K$ But: :$\displaystyle \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K = \bigcap \mathbb K_2$ and so: :$\displaystyle f \sqbrk {\bigcap \mathbb K_1} \subseteq \bigcap \mathbb K_2$ which means that: :$f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$ as we wanted to show. {{qed|lemma}} === Sufficient Condition === Suppose $f$ is not continuous. From Continuity Defined from Closed Sets, $\exists B \subseteq X_2$ which is closed in $T_2$ such that $f^{-1} \sqbrk B$ is not closed in $T_1$. By Image of Preimage under Mapping, we have that: :$f \sqbrk {f^{-1} \sqbrk B} \subseteq B$ So from Topological Closure of Subset is Subset of Topological Closure: :$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B^-$ From Closed Set Equals its Closure we have that $B^- = B$. Transitively, we get: :$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$ Because $f^{-1} \sqbrk B$ is not closed in $T_1$, we have that: :$f^{-1} \sqbrk B \subsetneq \paren {f^{-1} \sqbrk B}^-$ This means there exists an element $x \in \paren {f^{-1} \sqbrk B}^-$ such that $x \notin f^{-1} \sqbrk B$.  Therefore $\map f x \notin B$, but $\map f x \in f \sqbrk {\sqbrk {f^{-1} \sqbrk B}^-}$. From above, we had: :$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$ so there exists a set $A \subseteq X_1$, namely $A = f^{-1} \sqbrk B$, such that: :$f \sqbrk {A^-} \nsubseteq \paren {f \sqbrk A}^-$ {{qed}}