Continuous Closed Surjective Mapping is Quotient Mapping

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Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

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


Then $f$ is a quotient mapping.


Proof

Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.

For $X \subseteq S$, let $\relcomp S X$ denote the relative complement of $X$ is $S$.

By definition of closed set, $\relcomp {S_1} {f^{-1} \sqbrk U}$ is closed in $T_1$.

By definition of closed mapping, $f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} }$ is closed in $T_2$.

By definition of closed set, $\relcomp {S_2} {f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} } }$ is open in $T_2$.

Then:

\(\ds \relcomp {S_2} {f \sqbrk {\relcomp {S_1} {f^{-1} \sqbrk U} } }\) \(=\) \(\ds \relcomp {S_2} {f \sqbrk {f^{-1} \sqbrk {\relcomp {S_2} U} } }\) Complement of Preimage equals Preimage of Complement
\(\ds \) \(=\) \(\ds \relcomp {S_2} {\relcomp {S_2} U}\) Image of Preimage under Mapping: Corollary
\(\ds \) \(=\) \(\ds U\) Relative Complement of Relative Complement


By definition of quotient mapping, it follows that $f$ is a quotient mapping.

$\blacksquare$


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