Quotients of Homeomorphic Spaces are Homeomorphic

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $X, Y$ be topological spaces.

Let $\phi : X \to Y$ be a homeomorphism from $X$ to $Y$.

Let:

$\RR_X \subseteq X \times X$
$\RR_Y \subseteq Y \times Y$

be equivalence relations on $X$ and $Y$, respectively.

Suppose that, for all $x, x' \in X$:

$\map {\RR_X} {x, x'} \iff \map {\RR_Y} {\map \phi x, \map \phi {x'}}$


Then:

$X / \RR_X \sim Y / \RR_Y$

where $\sim$ denotes homeomorphic spaces.


Proof

Let the mapping $\psi : X / \RR_X \to Y / \RR_Y$ be defined as:

$\map \psi {\eqclass x {\RR_X}} = \eqclass {\map \phi x} {\RR_Y}$

In order for $\psi$ to be well-defined, the image needs to be independent of the choice of representative $x$.

But, for $x, x' \in X$ such that $\map {\RR_X} {x, x'}$, we have:

$\map {\RR_Y} {\map \phi x, \map \phi {x'}}$

by hypothesis.

It follows that:

$\eqclass {\map \phi x} {\RR_Y} = \eqclass {\map \phi {x'}} {\RR_Y}$


The inverse mapping $\psi^{-1} : Y / \RR_Y \to X / \RR_X$ can be defined as:

$\map {\psi^{-1}} {\eqclass y {\RR_Y}} = \eqclass {\map {\phi^{-1}} y} {\RR_X}$

For the same reasons as before, $\psi^{-1}$ is well-defined.

Additionally, it is the inverse of $\psi$, since:

\(\ds \map {\psi^{-1} } {\map \psi {\eqclass x {\RR_X} } }\) \(=\) \(\ds \map {\psi^{-1} } {\eqclass {\map \psi x} {\RR_Y} }\) Definition of $\psi$
\(\ds \) \(=\) \(\ds \eqclass {\map {\phi^{-1} } {\map \psi x} } {\RR_X}\) Definition of $\psi^{-1}$
\(\ds \) \(=\) \(\ds \eqclass x {\RR_X}\) Definition of Inverse Mapping
\(\ds \map \psi {\map {\psi^{-1} } {\eqclass y {\RR_Y} } }\) \(=\) \(\ds \eqclass y {\RR_Y}\) Symmetrically


It remains to show that $\psi$ and $\psi^{-1}$ are continuous.

Consider an arbitrary open set $U \subseteq Y / \RR_Y$.

By definition of quotient topology, we have that:

$\map {q_{\RR_Y}^{-1}} U$

is open in $Y$, where $q_{\RR_Y}$ is the quotient mapping induced by $\RR_Y$.

Then, since $\phi$ is continuous:

$\map {\phi^{-1}} {\map {q_{\RR_Y}^{-1}} U}$

is open in $X$.

If we can show that:

$\map {\psi^{-1}} U = \map {\phi^{-1}} {\map {q_{\RR_Y}^{-1}} U}$

then, since $U$ was arbitrary, $\psi$ would be continuous.

We have:

\(\ds \eqclass x {\RR_X} \in \map {\psi^{-1} } U\) \(\iff\) \(\ds \exists \eqclass y {\RR_Y} \in U: \map {\psi^{-1} } {\eqclass y {\RR_Y} } = \eqclass x {\RR_X}\) Definition of Image of Set under Mapping
\(\ds \) \(\iff\) \(\ds \exists \eqclass y {\RR_Y} \in U: \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}\) Definition of $\psi^{-1}$
\(\ds \) \(\iff\) \(\ds \exists y \in Y: \map {q_{\RR_Y} } y \in U \land \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}\) Definition of Quotient Mapping
\(\ds \) \(\iff\) \(\ds \exists y \in \map {q_{\RR_Y}^{-1} } U: \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}\) Definition of Inverse Mapping
\(\ds \) \(\iff\) \(\ds \eqclass x {\RR_X} \in \map {\phi^{-1} } {\map {q_{\RR_Y}^{-1} } U}\)

Therefore, by the above remarks, $\psi$ is a continuous mapping.

By a precisely symmetric argument, it can be shown that $\psi^{-1}$ is likewise continuous.


Therefore, since $\psi$:

it is a homeomorphism by definition.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Quotients_of_Homeomorphic_Spaces_are_Homeomorphic&oldid=686119"