Homeomorphism iff Image of Closure equals Closure of Image

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

Let $f: T_1 \to T_2$ be a bijection.

Then $f$ is a homeomorphism :
 * $\forall H \subseteq T_1: f \sqbrk {H^-} = \paren {f \sqbrk H}^-$

where $H^-$ denotes the closure of $H$.

Proof
From Bijection iff Inverse is Bijection, $f$ is a bijection $f^{-1}$ is a bijection.

From Continuity Defined by Closure we have that $f: T_1 \to T_2$ is continuous :
 * $\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$

Similarly, $f^{-1}: T_2 \to T_1$ is continuous :
 * $\forall H \subseteq S_2: f^{-1} \sqbrk {H^-} \subseteq \paren {f^{-1} \sqbrk H}^-$

Thus:
 * $\forall H \subseteq S_1: f \sqbrk {\paren {f^{-1} \sqbrk H}^-} \subseteq \paren {f \sqbrk {f^{-1} \sqbrk H} }^-$

That is:
 * $\forall H \subseteq S_1: f \sqbrk {\paren {f^{-1} \sqbrk H}^-} \subseteq H^-$

Hence the result.