Equivalence of Definitions of Homeomorphisms between Topological Spaces

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Theorem

Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.

Let $f: T_\alpha \to T_\beta$ be a bijection.


The following definitions of the concept of homeomorphism are equivalent:

Definition 1

$f$ is a homeomorphism if and only if both $f$ and $f^{-1}$ are continuous.


Definition 2

$f$ is a homeomorphism if and only if:

$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$


Definition 3

$f$ is a homeomorphism if and only if $f$ is both an open mapping and a continuous mapping.


Definition 4

$f$ is a homeomorphism if and only if $f$ is both a closed mapping and a continuous mapping.


Proof

Definition 1 iff Definition 2

Let $f$ and $f^{-1}$ both be continuous.

As $f$ is continuous, then by definition:

$V \in \tau_\beta \implies f^{-1} \sqbrk V \in \tau_\alpha$

and as $f^{-1}$ is continuous, then by definition:

$U \in \tau_\alpha \implies \paren {f^{-1} }^{-1} \sqbrk U = f \sqbrk U \in \tau_\beta$

That is:

$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$


Conversely, suppose that:

$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$

By definition:

$U \in \tau_\alpha \implies f \sqbrk U \in \tau_\beta$ if and only if $f^{-1}$ is continuous.

Also by definition:

$f \sqbrk U \in \tau_\beta \implies U = f^{-1} \sqbrk {f \sqbrk U} \in \tau_\alpha$ if and only if $f$ is continuous.

That is, both $f$ and $f^{-1}$ are continuous.

$\Box$


Definition 1 iff Definition 3

Let $f$ and $f^{-1}$ both be continuous.

Then by Bijection is Open iff Inverse is Continuous, $f$ is open.

Thus $f$ is both open and continuous.


Conversely, let $f$ be both open and continuous.

Then also by Bijection is Open iff Inverse is Continuous, $f^{-1}$ is continuous.

Thus both $f$ and $f^{-1}$ are continuous.

$\Box$


Definition 3 iff Definition 4

Let $f$ be both open and continuous.

Then by Bijection is Open iff Closed it follows that $f$ is both closed and continuous.


Conversely, let $f$ be both both closed and continuous.

Then also by Bijection is Open iff Closed it follows that $f$ is both open and continuous.

$\blacksquare$

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