Identification Topology equals Quotient Topology on Induced Equivalence

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Theorem

Let $\struct {S_1, \tau_1}$ be a topological space.

Let $S_2$ be a set.

Let $f: S_1 \to S_2$ be a surjective mapping.

Let $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $\struct {S_1, \tau_1}$:

$\tau_2 = \set {V \subseteq S_2 : f^{-1} \sqbrk V \in \tau_1}$

Let $\RR_f \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $f$:

$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$

Let $q_{\RR_f}$ be the quotient mapping induced by $\RR_f$.

Let $\tau_{\RR_f}$ be the quotient topology on $S_1 / \RR_f$ by $q_{\RR_f}$:

$\tau_{\RR_f} := \set {U \subseteq S_1 / \RR_f: q_{\RR_f}^{-1} \sqbrk U \in \tau_1}$


Then:

$\struct{ S_2, \tau_2 }$ is homeomorphic to the quotient space $\struct{ S_1 / \RR_f, \tau_{\RR_f} }$.


Proof

Define a mapping $\tilde f: S_1 / \RR_f \to S_2$ by:

$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$
$\begin{xy} \xymatrix@L+2mu@+1em{ S_1 \ar[r]^*{f} \ar[rd]_*{ q_{\RR_f} } & S_2 \\ & S_1 / \RR_f \ar@{-->}[u]^*{\tilde f} }\end{xy}$

Then $\tilde f$ is well-defined, as for all $s' \in \eqclass s { \RR_f}$, we have:

$\map f {s'} = \map f s$

For any $s_1 , s_2 \in S_1$ with $\eqclass {s_1}{\RR_f} \ne \eqclass {s_2}{\RR_f}$, we have:

$\map f {s_1} \ne \map f {s_2}$

so:

$\map {\tilde f}{\eqclass {s_1}{ \RR_f} } \ne \map {\tilde f}{\eqclass {s_2}{ \RR_f} }$

It follows that $\tilde f$ is injective.

For all $t \in S_2$, we can find $s \in S_1$ such that $\map f s = t$, as $f$ is surjective.

Then $\map {\tilde f}{\eqclass s { \RR_f} } = t$, so $\tilde f$ is surjective.

Let $V \in \tau_2$.

As Identification Mapping is Continuous, it follows that:

$ f^{-1} \sqbrk V \in \tau_1$

Then:

\(\ds f^{-1} \sqbrk V\) \(=\) \(\ds \set {s \in S_1 : \map f s \in V}\)
\(\ds \) \(=\) \(\ds q_{\RR_f}^{-1} \sqbrk {\set {\eqclass s {\RR_f} \in S_1 / \RR_f : \map f s \in V} }\)


By definition of quotient topology, we have:

$\set {\eqclass s {\RR_f} \in S_1 / \RR_f : \map f s \in V} \in \tau_{\RR_f}$

By definition of continuity, $\tilde f$ is continuous.

Let $U \in \tau_{\RR_f}$.

By definition of quotient topology, we have:

$q_{\RR_f}^{-1} \sqbrk U \in \tau_1$

By definition of identification topology, we have:

$f \sqbrk {q_{\RR_f}^{-1} \sqbrk U } \in \tau_2$

Then:

\(\ds f \sqbrk {q_{\RR_f}^{-1} \sqbrk U}\) \(=\) \(\ds f \sqbrk {\set {s \in S_1 : \map {q_{\RR_f} } s \in U} }\)
\(\ds \) \(=\) \(\ds \set {t \in S_2 : \map f s = t, \eqclass s {\RR_f} \in U}\)
\(\ds \) \(=\) \(\ds \tilde f \sqbrk U\)
\(\ds \) \(=\) \(\ds \paren {\tilde f^{-1} }^{-1} \sqbrk U\)


By definition of continuity, $\tilde f^{-1}$ is continuous.

By definition of homeomorphism, $\tilde f$ is a homeomorphism between $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$.

$\blacksquare$


Sources

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