Theorem: Quotients of Homeomorphic Spaces are Homeomorphic

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:
{{begin-eqn}}
{{eqn | l = \map {\psi^{-1} } {\map \psi {\eqclass x {\RR_X} } }
      | r = \map {\psi^{-1} } {\eqclass {\map \psi x} {\RR_Y} }
      | c = Definition of $\psi$
}}
{{eqn | r = \eqclass {\map {\phi^{-1} } {\map \psi x} } {\RR_X}
      | c = Definition of $\psi^{-1}$
}}
{{eqn | r = \eqclass x {\RR_X}
      | c = {{Defof|Inverse Mapping}}
}}
{{eqn | l = \map \psi {\map {\psi^{-1} } {\eqclass y {\RR_Y} } }
      | r = \eqclass y {\RR_Y}
      | c = symmetrically
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \eqclass x {\RR_X} \in \map {\psi^{-1} } U
      | o = \iff
      | r = \exists \eqclass y {\RR_Y} \in U: \map {\psi^{-1} } {\eqclass y {\RR_Y} } = \eqclass x {\RR_X}
      | c = {{Defof|Image of Set under Mapping}}
}}
{{eqn | o = \iff
      | r = \exists \eqclass y {\RR_Y} \in U: \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}
      | c = Definition of $\psi^{-1}$
}}
{{eqn | o = \iff
      | r = \exists y \in Y: \map {q_{\RR_Y} } y \in U \land \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}
      | c = {{Defof|Quotient Mapping}}
}}
{{eqn | o = \iff
      | r = \exists y \in \map {q_{\RR_Y}^{-1} } U: \eqclass {\map {\phi^{-1} } y} {\RR_X} = \eqclass x {\RR_X}
      | c = {{Defof|Inverse Mapping}}
}}
{{eqn | o = \iff
      | r = \eqclass x {\RR_X} \in \map {\phi^{-1} } {\map {q_{\RR_Y}^{-1} } U}
}}
{{end-eqn}}
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 $\psi$:
* is a bijection
* is continuous
* has a continuous inverse
it is a homeomorphism by definition.
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Quotient Spaces (Topology)
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