Closed Image of Closure of Set under Continuous Mapping equals Closure of Image

Theorem
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $H \subseteq S_1$ be a subset of $S_1$.

Let $\map \cl H$ denote the closure of $H$.

Let $f: T_1 \to T_2$ be a continuous mapping.

Let $f \sqbrk {\map \cl H}$ be closed in $T_2$.

Then:
 * $f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$

Proof
By Continuity Defined by Closure:
 * $f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$

The proof follows by definition of set equality.