Continuous Closed Surjective Mapping is Quotient Mapping

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:

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