Homeomorphism iff Image of Closure equals Closure of Image

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Definition

Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

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


Then $f$ is a homeomorphism if and only if:

$\forall H \subseteq T_1: f \left({H^-}\right) = \left({f \left({H}\right)}\right)^-$

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


Proof

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

From Continuity Defined by Closure we have that $f: T_1 \to T_2$ is continuous if and only if:

$\forall H \subseteq S_1: f \left({H^-}\right) \subseteq \left({f \left({H}\right)}\right)^-$

Similarly, $f^{-1}: T_2 \to T_1$ is continuous if and only if:

$\forall H \subseteq S_2: f^{-1} \left({H^-}\right) \subseteq \left({f^{-1} \left({H}\right)}\right)^-$

Thus:

$\forall H \subseteq S_1: f \left({\left({f^{-1} \left({H}\right)}\right)^-}\right) \subseteq \left({f \left({f^{-1} \left({H}\right)}\right)}\right)^-$

That is:

$\forall H \subseteq S_1: f \left({\left({f^{-1} \left({H}\right)}\right)^-}\right) \subseteq H^-$

Hence the result.

$\blacksquare$



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